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COPYRIGHT DEPOSn^ 



THE ELEMENTS OF MECHANICS 
OF MATERIALS 

A TEXT FOR 
STUDENTS IN ENGINEERING COURSES 






BY 



C. E. HOUGHTON, A.B., M.M.E. 

ASSOCIATE PROFESSOR OF MECHANICAL ENGINEERING 
NEW YORK UNIVERSITY 



ILLUSTRATED 



SECOND EDITION 
REVISED AND ENLARGED 



NEW YORK 

D. VAN NOSTRAND COMPANY 

1915 



J 



-rK^?^ 






Copyright, 1909, 
By D. van NOSTRAND COMPANY. 



Copyright, 1915, 
By D. van NOSTRAND COMPANY. 



THE SCIENTIFIC PRESS 

ROBERT DRUMMOND AND COMPANY 

BROOKLYN. N. Y. 



QGT-4I9I5 



PREFACE 

This is not a treatise on the Mechanics of Materials. 
The efforts of Merriman, Burr, Lanza, and others cover 
the field so thoroughly that there is no present need of 
such a work. 

It is designed to be an elementary text-book for students 
in the engineering courses in colleges and universities, 
where the time allotted to the subject does not exceed 
three or four recitations per week, for one half year, and 
where the course is preceded by college courses in mathe- 
matics, through integral calculus, mechanics, and physics. 

The extreme mathematical treatment of the subject has 
been avoided, but where the use of higher mathematics 
leads to clearness they have been freely used. 

As it is intended as a text-book, the general cases are 
discussed fully, leaving the student to derive the formulas 
for special cases as part of the regular problem work. 

At the end of each chapter there are review questions 
covering the more important parts of the subjects dis- 
cussed and problems illustrating the s^^me. The solution 
of one problem of each type has been given to show the 
application of the general formulas. 

The appendix contains tables giving the values of the 
engineering constants of materials and the formulas com- 
monly used in design, in addition to the tables usually 
found in books of this character. 

The notation has been made uniform with that of Mer- 
riman's works, so that his more complete treatise on the 
subject may be conveniently used as a reference book. 

Nbw York, January, 1909. 

lU 



TABLE OF CONTENTS 

CHAPTER I 
Applied Mechanics 

AJBTIOLE PAGE 

1. Forces in structures 1 

2. Axial forces 3 

3. A bar . 3 

4. Internal forces 3 

6. Tensile or compressive stresses 4 

6. Unit stress 4 

7. Maximum tensile or compressive stresses .... 5 

8. Shearing stresses , . 5 

9. External and internal forces 6 

10. Deformation of elastic bodies 7 

11. Unit deformations . . . . ... . . 7 

12. Modulus or coefficient of elasticity . . . . .8 

13. The elastic limit 8 

14. Ultimate strength 9 

15. Resilience 10 

16. Ductility. 11 

17. Elastic resilience .12 

18. Use of formulas . . 12 

19. Constants of materials 13 

20. Units 14 

21. Working stresses ; factors of safety 14 

22. Accuracy of calculations 15 

Examination questions 16 

Problems 18 



CHAPTER II 

Applications 

23. Bars of uniform strength , .22 

24. Thin pipes, cylinders, and spheres 25 



VI 



TABLE OF CONTENTS 



25. Thick pipes .... 

26. Riveted joints . . • 

27. Tension in plates . 

28. Shear on rivets 

29. Compression on rivets or plates 

30. General case of a riveted joint 

31. Kinds of riveted joints . 

32. Efficiency of a riveted joint . 

33. Stresses due to change of temperature 
Problems 



PAOK 

^7 
30 
30 
32 
33 
33 
35 
37 
39 
40 



CHAPTER III 

Beams 

34. Kinds of beams 46 

35. Reactions at the supports . . . . . . .47 

36. Uniform and concentrated loads 47 

37. Vertical shear . . .49 

38. Bending moment . .50 

39. Resisting shear 51 

40. Resisting moment .52 

41. Use of formula 55 

42. Shear and moment diagrams 58 

43. Shear diagrams 59 

44. Moment diagrams 60 

45. The relation between the vertical shear and bending moment 63 

46. Relative strengths of simple and cantilever beams . . 64 

47. Overhanging beams 65 

48. Beams of uniform strength 66 

49. Moving loads . . 67 

50. Use of formula 69 

51. Examination questions 70 

Problems 73 



CHAPTER IV 
Torsion 



52. Derivation of formula 

53. Modulus of section 

54. Square sections 



83 
86 
86 



TABLE OF CONTENTS 



Yll 



ARTICLE 

55. Illustrations . . . . 

56. Twist of shafts 

57. Relative strengths and stiffness 

58. Horse power of shafts . 

59. Shaft couplings 

60. Modulus of rupture for torsion 

61. Helical springs 
Examination questions . 
Problems .... 



PAGE 

87 



89 
90 
92 
93 
94 
95 



CHAPTER V 

The Elastic Curve 

62. Definition .99 

63. The equation of the elastic curve 99 

64. Deflection of beams 102 

65. Fixed or restrained beams . 105 

66. Beams fixed at both ends 107 

67. Continuous beams . . . 109 

Examination questions 114 

Problems . . . ^ 116 



CHAPTER VI 

Long Columns 

68. Stresses in long columns 

69. Euler's formula for long columns . 

70. Columns with round or pin ends . 

71. Columns with square, flat, or fixed ends 

72. Columns with round and square ends 

73. Rankine's formula for long columns 

74. Applications 

Examination questions . 
Problems 



123 
124 
125 
126 
128 
131 
135 
187 
139 



CHAPTER Vn 

Combined Stresses 

75. Stresses due to force 

76. Tension or compression combined with bending 



143 

143 



viii TABLE OF CONTENTS 

ARTICLE PAGE 

77. Roof rafters . 147 

78. Eccentric axial loads . , , . . , , . 149 

79. Shear and axial stress ..,,,,,. 149 

80. Maximum stresses • . , . 150 

81. Horizontal shear in beams 152 

82. Maximum stresses in beams . 156 

Examination questions 158 

Problems 160 

CHAPTER Vm 

Compound Bars and Beams 

83. Definition 163 

84. Compound columns, alternate layers 163 

85. Compound columns, longitudinal layers .... 163 

86. Compound beams . 165 

87. Reenforced concrete beams ....... 166 

88. Straight line fornmla for reenforced concrete beams . . 166 

Examination questions 173 

Problems . . 174 

Tables, Explanation of 176 

Table 1. Notation . . . . • 176 

Table 2. Fundamental formulas 178 

Table 3. Derived formulas 179 

Table 4. Properties of beams 182 

Table 5. Average constants of materials 183 

Table 6. Properties of sections 184 



MECHANICS OF MATERIALS 



CHAPTER I 
APPLIED MECHANICS 

Article 1. Forces in Structures. 

One of the problems that confronts the engineer called 
upon to design any machine or structure is to so propor- 
tion the various parts that they will resist the forces that 
act on them. 

To do this, he must apply the laws of mechanics to the 
forces to be resisted, and study the action of the materials 
under the same forces. 

This application of mechanics may be termed Applied 
Mechanics or the Mechanics of Materials. 

If we consider any structure or any member of the 
structure to. be at rest, according to the laws of mechanics 
the forces that act on the structure or member must be in 
equilibrium. The various parts of a machine often have 
relative motion, but by introducing a force equal and op- 
posite to the force Avhich produces that motion, the forces 
that act on the member of a machine may also be treated 
as a system of forces in equilibrium. 

In the more extended treatment of this subject the 
forces are taken as acting in different planes. The simpler 
theory that treats the forces as coplanar is the one that 
will be used here. 

1 



2 MECHANICS OF MATERIALS 

Since each member must be designed separately, if the 
forces that act on any member are determined, we will find 
that the forces may be resolved into : 

Forces tending to lengthen the member, 
Forces tending to shorten the member, 
Forces tending to bend the member, 
Forces tending to shear the member. 

Imagine any body, acted on by a system of forces in 
equilibrium, denoting the sum of the components of the 
forces parallel to some line as the -X" forces, and those per- 
pendicular to the same line as the ]r forces; the sum of the 
X forces is zero and the sum of the Y forces is zero. If 
we take as the line of reference the axis of the body pass- 
ing through the center of gravity of the body, the result- 
ant of the X forces will be a couple, unless the lines of 
action of the forces of the couple coincide with the axis. 

Each force of this couple may be replaced by a single 
force of equal magnitude, acting in the line of the axis and 
a couple whose moment is the moment of the force about a 
point in the axis. The X forces acting in the line of the 
axis will tend to either lengthen or shorten the member 
in the line of the axis, and the couple as well as the Y 
forces will tend to bend the member. 

If the member be cut by a plane perpendicular to the 
axis, the Y forces on either side of the section will be 
opposite in sign, and in general tend to slide one part of 
the member, relative to the other, along the plane of the 
section. Any of the resultants may be zero, and in that 
case there would be no tendency to deformation in that 
line. 

The force acting on any member is always transmitted 
by a surface of finite area, but by considering that each 



APPLIED MECHANICS 



elementary area of the surfaces in contact transmits the 
same amount of pressure, we may use the resultant of 
these elementary pressures passing through the center of 
gravity of the areas in contact, as the force applied. 

Art. 2. Axial Forces. 

When the lines of action of the acting forces lie in 
the axis of the body, passing through the centers of 
gravity of all sections perpendicular to the axis, the forces 

are called Axial 
Forces^ and their 
effect is to 
either lengthen 
or shorten the 
member. 



V 



\r 



a 



\^ 




Fig. 4. 



a 

Art. 3. A Bar. 

The member 
may have any 
shape whatever, 
JC the simplest be- 
ing a prismatic 
or cylindrical 
form, where any section perpendicular 
to the axis has the same shape and area. 
This form will be called a Bar. 

Art. 4. Internal Forces or Stresses. 

Let Fig. 4 represent a bar under the 
action of the axial forces P and P. 
Suppose the bar to be cut by any plane 
through the axis into the segments a 
and 6, and consider the segment a. 



4 MECHANICS OF MATERIALS 

This segment is acted on by the external force P, and as a 
part of the whole bar it is in equilibrium ; hence there must 
be forces acting in the plane section X-X^ whose resultant 
acts in the same line, and is equal and opposite to the force 
P. As the same is true of the segment 5, the internal 
forces, called stresses, acting between a and 5, hold the 
segments a and h in equilibrium against the external forces. 
Therefore, in any plane section of the bar there exists 
a pair of equal and opposite forces or stresses, each of 
which are induced by and resist the external forces. In 
general, the stresses may be resolved into components 
parallel and perpendicular to the plane section. The 
components parallel to the plane of the section prevent 
the sliding of the segments along the plane, and are 
termed Shearing Stresses, while those perpendicular to the 
plane are called either Tensile or Compressive Stresses, 
depending on whether they tend to extend or compress 
the particles on which they act. 

Art. 5. Tensile or Compressive Stresses. 

When the external forces are axial, and the section per- 
pendicular to the axis, the stresses can have no component 
parallel to the plane of the section; hence axial forces can 
produce only tensile or compressive stresses in planes per- 
pendicular to the axis. The plane dividing the bar into 
the segments a and b was any plane ; hence the reasoning 
holds true for all such planes, and there are only tensile or 
compressive stresses equal to the external force P, in all 
sections perpendicular to the axis. 

Art. 6. Unit Stress. 

Since the force P is the resultant of all the equal unit 
pressures on the areas in contact, it is reasonable to assume 



APPLIED MECHANICS 5 

that the stresses on each unit of area of the plane section 
are also equal, and if S equals the sum of the stresses 
acting on each unit of area of the plane section, and A is 
the area of that section, then 

F = AS. (a) 

S is termed the unit stress, and is the resisting force per 
square unit of area ; hence S must be expressed in the 
same units as P and A. 

Art. 7. Maximum Tensile or Compressive Stresses. 

When the cutting plane is not perpendicular to the 
axis, the resultant stress may be resolved into components 
parallel and perpendicular to the plane of the section, 
those perpendicular being either tensile or compressive 
stresses, while those parallel are shearing stresses. 

As neither component can equal the resultant, it is evi- 
dent that the maximum tensile or compressive stresses 
v^ill be found in a section perpendicular to the axis. In 
such a section there are no shearing stresses, and when 
the bar has. a uniform section area A, the formula (tx) will 
determine the maximum tensile or compressive unit stress 
induced by the axial force P. If the areas of all sections 
perpendicular to the axis are not equal, the greatest unit 
stress will be found where the section area is the least, 
and the value of A to be used in formula (a) is the area 
of the least section. 

Art. 8. Shearing Stresses. 

When the external forces act in adjacent parallel lines, 
since the stresses can have no component perpendicular to 
the line of action of the forces, the stress in a section 
parallel to the line of action of the forces must be a 



6 



MECHANICS OF MATERIALS 



Shearing Stress^ as the external forces tend to slide the two 
sections of the bar along the plane of the section. (Fig. 8.) 
Assuming that the stress is uniformly distributed over 
the section, and that Sg is the unit stress in shear, then 
formula {a) P = AS^^ where A is the area of the section 




5 



Fig. 8. 

parallel to the line of the forces and P the forces pro- 
ducing the shear, will always give the relation between 
the external forces and the maximum unit shearing stress 
in the section. 

Art. 9. External and Internal Forces. 

The external forces on any member of a machine or 
structure are the weights or loads that member has to 
support and the pressure it receives from the adjacent 
members, while the internal forces are those that transmit 
the external force from element to element through the 
member. These latter forces are stresses, and the inter- 
nal force per unit of area is the Unit Stress^ and will be 
designated by the letter aS', with a subscript ^, c, or s, as 
the stress is tension, compression, or shear. 

The formula P = AS is a general one and applies to all 
cases where . the stress is uniformly distributed over the 
area of any plane section A^ and S, the kind of unit 
stress. 



APPLIED MECHANICS 7 

Art. 10. Deformation of Elastic Bodies. 

In the study of mechanics, the forces were assumed to 
act on rigid bodies; that is, on bodies whose shape was not 
altered by the application of the force. As there are no 
rigid bodies in nature, every force applied produces some 
deformation or change of shape. This fact, however, does 
not prevent the application of the laws of mechanics to 
elastic bodies under the action of force after the deforma- 
tion has taken place, since equilibrium must exist at that 
time. 

Art. 11. Unit Deformations. 

Consider a bar, I units in length, and A square units in 
section, under the action of an axial force P. From equa- 

P 

tion (a), S = --, S\^ constant, since P and A are constant 

A 

for all sections perpendicular to the axis, or each square 

unit of every section is acted on by a force S. 

Suppose the bar to be divided into bars, each one unit 
in section and I units long, then each of these bars is acted 
on by a force S. The change in the length ?, that may 
take place under the force S^ being e, since all particles 
are under the same force, the change in the length must 
be equal to the length Z, multiplied by the unit load S^ 
and some number which depends on the nature of the 

material. Calling this number -— , the value of e must be 

e =—. This may be written - = — . Letting r = e equal 
-tj I IE I 

the change in the length of a bar one unit in length, or the 
unit deformation, it follows that e = — or 

■p _S_ unit stress ^7 >. 

€ unit deformation 



8 MECHANICS OF IVLATERIALS 

Art. 12. Modulus or Coefficient of Elasticity. 

If the value of S does not exceed a critical value which 
varies for different materials, U will be constant for all 
values of S, and this constant is called tlie Modulus or Co- 
efficient of Elasticity. This constant is the value of tlie 
ratio of the unit stress to the unit deformation, as may be 
seen by the inspection of formula (6). 

PI P 

Equation (5) may be written E = — -, since S = ~ and 

e = -, and as A and I are constants for any bar under axial 
forces, E will be constant when e varies directly with P. 

Art. 13. Elastic Limit. 

If a bar length I is subjected to a small axial force P, it 
is observed that the length has changed a certain small 
amount. If Pg is twice P, experiment shows that the 
change in length is twice that due to P, and if P^ is n 
times P, the change in the length is n times that due to 
P, provided that P^ does not exceed a certain limiting 
value which varies for each material and bar. That is, 
within that limit the change in length of any bar is pro- 
portional to the external force applied. If P^ is the 
limiting value for a given bar, then the corresponding 
value of S as derived from formula {a) must be the 
limiting value of the unit stress. This value of S^ being 
a unit stress, is independent of the dimensions of the 
bar and depends only on the material of the bar, is 
called the Elastic Limit of the material or the limit of 
elasticity. 

The elastic limit of a material may be defined as the 
unit stress for which the deformations cease to be propor- 
tional to the applied force. 



APPLIED MECHANICS 9 

To determine this value of S for any material, axial loads 
are applied to a bar of the material, the loads being applied 
in small equal increments, and the change in length due to 
each increment of load measured. For any total load less 
than the load producing a stress equal to the elastic limit, 
the last increment of load should produce the same change 
in length as any previous increment. Therefore, when an 
increase in the change in length for any equal increment 
of load is noted, the elastic limit has been passed. It will 
be noticed that the exact value of elastic limit depends on 
the accuracy with which the loads and especially the de- 
formations are measured. 

It has also been observed that if the stress in any bar is 
less than the elastic limit, the bar will return to its origi- 
nal length when the load is removed, and if the stress is 
slightly above the elastic limit, there will be some perma- 
nent change in the length of the bar. The unit stress at 
which this yielding takes place is called the Yield Point 
or the Commercial Elastic Limit. The latter term comes 
from the commercial practice of determining the elastic 
limit by the drop of the beam in the testing machine. 

While the yield point or commercial elastic limit 
is from 3 to 5 % higher than the true elastic limit, the ease 
with which the latter value may be determined and the 
fact that the allowable value of aS' for any engineering 
structure rarely exceeds one half of the elastic limit, com- 
bine to make it the one in general use. 

Art. 14. Ultimate Strength. 

After the elastic limit is reached, the change in length 
increases more and more rapidly as the loads are increased, 
and finally a load is reached that causes the bar to rupture. 



10 MECHANICS OF MATERIALS 

If P is the load that causes rupture and A is the original 
area of the bar, then the value of S obtained from formula 
(a) is called the Ultimate Strength of the material. 

Art. 15. Resilience. 

If a bar of wrought iron, whose section area is A^ and 
the length I as measured between two punch marks on 
the bar, is placed in a testing machine, the tensile loads 
and their corresponding extensions being measured and 
plotted to scale on section paper, the result would be a 
diagram similar to that in Fig. 15. 









§ 









b 




a 
/T — 


^.^^^^^^'^^ 




/ 
/I 
/ 1 

1 1 


1 







y f Change in /en gth, or unit- deformation. 



Fig. 15. 

Since P = AS^ the ordinates representing loads may, 
by a change of scale, represent unit stresses, and the ab- 
scissa representing total elongations may also represent 
unit elongations. Such a diagram is called a Stress-strain 
diagram. The use of the word "strain" gives it the 
meaning of "unit deformation." Authorities do not agree 
on the use of the word, some giving it the meaning as 
above, while others use it to mean load. On account of 



APPLIED MECHANICS 11 

this ambiguity, the term "unit deformation" will be used 
in its place. The unit stress for any load is obtained by 
dividing the load by the original area, and the unit elon- 
gations by dividing the elongation for any load by the 
original length I, 

The point a on the curve is the elastic limit, and b 
is the ultimate strength of the material. After a load 
corresponding to h is reached, the bar begins to reduce at 
some point very rapidly, finally breaking at a load less 
than the load at h. This load is called the Breaking Load^ 
and has but little significance in engineering work. 

Since one coordinate represents force, and the other 
space, the area o ah c e, when measured in the proper 
units, is the work done in breaking a bar of unit volume. 
If we define the Ultimate Resilience as the work done in 
breaking a bar of unit volume, the area represents that 
quantity. 

Art. 16. Ductility. 

As the whole curve is rarely ever determined, the Duc- 
tility^ a term that is defined by its method of calculation 
and is proportional to the ultimate resilience, is generally 
used in its place. 

The ductility of any material is calculated as follows: 
After the bar has been broken by a tensile load, the pieces 
are removed from the testing machine and the broken ends 
placed together. The distance between tlie original punch 
marks, being ^, has now increased to Z+^; then -t. is called 

the Ductility^ and is usually stated as a percentage of the 
original length. 



12 MECHANICS OF MATERIALS 

Art. 17. Elastic Resilience. 

The area oaf is easily calculated, as it is the area of 
a triangle, and when measured in work units is called the 
Elastic Resilience. This work is evidently 

the unit stress at a x the unit deformation at a 

2 

Calling this value Ar, it is the elastic resilience of the 
material, or the work done in raising the unit stress in a 
bar of unit volume from to the elastic limit. 

The same reasoning is true for any stress less than the 
elastic limit, giving a method of calculating the work 
necessary to produce any given stress less than the elastic 
limit in a bar of unit volume. If S is any stress less than 
the elastic limit, and e the unit deformation at that stress, 
then ^ = J aS'c is the work done per unit of volume, and 
the work done on any bar whose volume is F' is 1^ 
times that quantity. Taking the work done on any 

bar as ^ Se x V, substituting for S its value --, for e, - 

and for F", Al, the expression for the work done on any 
bar reduces to K = ^ Pe which is in terms of the total 
load and deformation, and is true when P is less than the 
load, causing a stress within the elastic limit. 

Art. 18. Use of Formulas. 

Whenever any stress can be assumed to be uniformly 
distributed over the area of any section of a bar, formula 
(a), P = AS, gives the relation between the load and the 
stresses on that area, and any two of these quantities 
involved being given, the other may be found. 

The relation between the load and the deformation for 



APPLIED MECHANICS 13 

>S PI 
such cases is given by U = — = — — , which is true when 

e Ae 

aS' = — is less than the elastic limit, and is applicable to 
.A. 

all problems involving the deformations of a bar under 
axial loads. 

Also k = ^ Se and K = ^ Pe can be used under the 
same conditions when the data for the problem include a 
consideration of the work done in deforming a bar. When 
the load is axial and the stress may be taken as uniformly 
distributed over the area of any section, these three equa- 
tions furnish the means for the design and investigations 
of the strength of all members of a structure or machine. 

Writing (a) aS' = ^, (&) ^=-, and A; = 1 Se, it will be 

noticed that the equations are simply the algebraic ex- 
pressions of the definitions of Unit Stress, Modulus of 
Elasticity, and the Unit Resilience, making it easier to 
remember either the formula or the definition. 

Art. 19. Constants of Material. 

These three equations make use of the following con- 
stants of materials: 

Elastic Limit, 

Ultimate Strength, 

Modulus of Elasticity, and 

Modulus of Elastic Resilience, 
all of which have to be determined by experimental work 
on bars of the various materials. As this work cannot be 
done for each problem, a table containing average values 
of these constants for the more common materials will be 
found in the Appendix, and the question of the units in 
which each is expressed becomes important. 



14 MECHANICS OF MATERIALS 

Art. 20. Units. 

In American practice, the linear unit is the inch ; the 
square unit, the square inch, and the unit of weight, the 
pound. 

In the tables, the values of E. L., U., and E., are given 
in pounds / square inch and k in inch pounds. Therefore 
when the solution of any problem requires the use of any 
of these constants, all of the quantities involving weights 
or loads must be in pounds, and those involving linear 
or square measure, in inches. 

Art. 21. Working Stresses ; Factors of Safety. 

No material is entirely free from flaws and imperfections, 
which tend to diminish the area that is effective in resist- 
ing the external force, and in no case should the stress in 
any member be greater than the elastic limit, as such a 
stress would cause some permanent deformation. Sud- 
denly applied loads, shocks, and loads producing alternate 

tension and compression, all produce stresses that are 

p 

greater than the value of S = — , which is the value of S 

"" A 

for the same loads gradually and steadily applied. When 
any of these conditions occur, they tend to reduce the 
allowable or safe value of S to be used in the equation 
P = AS. Therefore, if U is the ultimate strength of the 
material, the allowable value of the unit stress can be found 
by dividing the ultimate strength by some number. Calling 

this number the Factor of Safety, F, S = —is tlie value of 

F 

fS to be used in connection with the formula F = AS, and 

is termed the Safe or Working Stress. 

The factor F depends — 



APPLIED MECHANICS 15 

1. On the reliability of the material ; that is, the lia- 
bility of flaws or imperfections that may reduce the effect- 
ive area of the section. 

2. On the way in which the loads are applied. 

The first part has been termed a factor of ignorance, 
while the second may be determined more or less accurately 
from theoretical considerations. 

The factors of safety as given in the tables in the Appen- 
dix are those to be used in the solution of the problems 
given in this book, and it must be remembered that the 
values are only approximate. 

The safety of any structure calling for a large factor, 
while the consideration of cost always demands the 
smallest one, the final choice of the factor of safety to be 
used in any given case must be largely a question of 
engineering judgment. In some cases, as in buildings, 
the allowable stress under the various kinds of loadings 
is a part of the building laws, and the engineer has to 
conform to the law\ 

For steady loads and reliable material the smallest fac- 
tor in general use is about four. 

On account of the danger of permanent injury to the 
material, no stress should exceed the elastic limit ; hence, it 
would seem better engineering to base the factor of safety 
on the elastic limit rather than on the ultimate strength, 
but such practice is not general in engineering work. 

Art. 22. Accuracy of Calculations. 

As the use of a factor of safety of four will result in an 
area of section about twice what it would have been had 
the allowable stress been equal to the elastic limit, and the 
values of E. L., U., and E., being determined by experi- 



16 MECHANICS OF MATERIALS 

ment, are liable to an error of from 3 to 5 %, there is no 
necessity for absolute accuracy in the calculations for the 
design of the different parts of a structure. 

Students are urged to use a slide rule in the numerical 
solution of the problems given in this text, not only to 
save time during the college course, but because they will 
find the use of a slide rule almost necessary in the prac-^ 
tice of their profession. The slide rule should always 
give results that are not in error more than 1, or at the 
outside 2 %, which is close enough for the greater part of 
engineering calculations. 

They are warned that it is the significant figures in a 
number that is to be used as a factor, and not the decimal 
point, that is of importance. If in one case the area was 
given as 12,500 sq. in. and as .00125 sq. in. in another, 
the figure 5 is of equal importance in each case. The 
use of the latter area as .0012 sq. in. will result in an error 
of 4 9^ in the value of the stress. 

EXAMINATION QUESTIONS 

1. There is no relative motion between the different 
parts of a bridge, therefore each part must be in equi- 
librium. How is it that the same laws may be applied 
to machine parts, which we know have relative motion ? 

2. When may forces be considered as Axial ? 

3. Define the term "Bar" as used in the text. 

4. What is a stress ? If there are stresses in every 
section of a bar, why is it that there is no relative motion 
between the different parts ? 

5. When may stresses be termed Tensile or Compres- 
sive Stresses ? 

6. What is a Unit Stress ? 



APPLIED MECHANICS 17 

7. If a member of any structure varies in size at dif- 
ferent parts of its length, how can you find the maximum 
tensile or compressive unit stress due to an axial load ? 

8. Give some examples of forces that produce tensile, 
compressive, and shearing stresses. 

9. What are the external forces for any member of 
a structure ? What are the internal forces ? 

10. Show that P = AS, give the units involved, and the 
limits of use for the formula. 

11. The science of mechanics is based on the action of 
forces on rigid bodies. Why is it that the same laws may 
be applied to forces acting on elastic bodies ? 

12. What is meant by the expression, Unit Deformation ? 

13. If the modulus of elasticity for steel is 30,000,000 
and for wrought iron is 25,000,000, and one bar of each is 
the same size and carries the same tensile load, which bar 
will stretch the most? 

14. The value of U may be termed a measure of the 
rigidity of a material. Why ? 

15. Define Elastic Limit and Ultimate Strength. 

16. Why does the Commercial Elastic Limit or Yield 
Point differ. from the true elastic limit? 

17. What is a Stress-strain diagram ? 

18. If a load corresponding to the ultimate strength of 
the material is placed on a bar, why is not that load 
called the Breaking Load ? Article 14 says otherwise. 
Explain. 

19. What is meant by the term Ductility ? 

20. Define Elastic Resilience. 

21. State the formulas for calculating the elastic resili- 
ence and modulus of elasticity ; give the units involved and 
the limits of use for each formula. 



18 MECHANICS OF MATERIALS 

22. What is a factor of safety ? A working stress ? 
What is the difference between a working stress and a safe 
stress ? What do you understand by a safe load ? 

23. Show why, if the calculations for the stresses in 
any member are not in error more than 2%, they are 
substantially correct. 

PROBLEMS 

1. A square steel bar 2x2 in. in section and 4 ft. 
long, carries a tensile load of 60,000 lb. Required the 
unit tensile stress. 

Solution. The relation between the load, area, and unit stress, 
when the load is axial, is always given hj P = AS, hence substituting 
for F and A, from the data given in the problem, 

60,000 = 4: X S, or S = 15,000 Ib./sq. in. 

2. A round wooden column, 16 in. in diameter and 
12 ft. 6 in. long, supports a load of 20 tons. Required 
the unit stress. 

3. What is the value of the maximum tensile load the 
bar in problem i will carry ? 

4. A wrought iron bar is 2 in. in diameter and 5 ft. 
long. What tensile load may be carried if the unit stress 
does not exceed 10,000 Ib./sq. in. ? 

5. What is the maximum tensile load the bar in prob- 
lem 4 will support ? 

6. A square cast iron column is hollow, 10 x 10 in. 
on the outside and 8x8 in. on the inside. Required the 
maximum compressive load that may be carried. 

7. In problem 6, keeping the outside dimensions the 
same, required the inside dimensions if the load is 360,000 
lb. and the unit stress is 10,000 Ib./sq. in. 

8. A punch is 1 in. in diameter. Required the prob- 
able pressure necessary to force the punch through a 
steel plate, ^ in. thick. 



APPLIED MECHANICS 19 

Solution. In order to force the punch through the plate, the 
unit shearing stress on an area equal to the cylindrical surface 
of the punched hole must be the ultimate shearing stress of the 
material ; hence, 

P = 7r(^«x»S' = 7rxlx^x 50,000 = 80,000 lb. approximately. 

9. Using a punch 1 in. in diameter and an available 
force of 80,000 lb., what is the thickest wrought iron that 
can be punched ? 

10. An iron casting is bolted to the floor by four 
one-inch bolts, and a force tends to slide the casting 
along the floor. Neglecting friction, what is the probable 
magnitude of the force when the unit shearing stress in 
the bolts is 10,000 lb. / sq. in. ? 

11. If the force in problem 10 was 24,000 lb., select 
four standard bolts, so that the unit shearing stress will 
not exceed 10,000 lb. / sq. in. 

12. If steel costs five cents per pound and wrought 
iron four cents, which will it be cheaper to use to carry a 
tensile load if the same factor of safety is used in each 
case ? 

Note. Assume the weights per cubic foot are the same for each ; 
then the weights in each case will be proportional to the areas of the 
sections. 

13. With wrought iron at four cents per pound, and 
other conditions the same as in 12, how much can you 
afford to pay for steel ? 

14. Required the probable elongation of the bar in 
problem 1. 

Solution. The relation between the elongation and an axial load 

PI 

is given by ^ — — , and substituting the data as given in the prob- 
Ae 

lem, 30,000,000 = ^Q^QQQ x ^^ , ... g := .024 in. 

4e 



20 MECHANICS OF MATERIALS 

15. How much work is done by the force in prob- 
lem 1 ? 

16. How much work is done by the force in prob- 
lem 4 ? 

17. A concrete pier 3 ft. by 4 ft. in area carries a load 
of 300 tons. Required the unit stress. 

18. A brick pier carries the same load with a unit 
stress of 18 tons / sq. ft. Required the area of the sec- 
tion. 

19. The thickness of the head of a standard bolt is 
approximately equal to the diameter of the bolt. Com- 
pare the unit tensile stress in the bolt with the unit shear- 
ing stress in the head. 

20. A standard steel bolt IJ in. in diameter supports 

a tensile load of 9800 lb. Required the factors of safety 

for the tensile and shearing stresses. 

(The least area to resist tension is at the root of the thread. See 
tables.) 

21. If the bar in problem l was 4 ft. 2 in. long and 
the elongation was .025 in., required the modulus of 
elasticity. 

22. If the modulus of elasticity of wood is 1,500,000, 
required the shortening of the column in problem 2. 

23. A steel bar 1 in. in diameter has two punch 
marks 8 in. apart marked on it. The bar is placed in 
a testing machine and it is found that there is a rapid 
change in the rate of elongation, when the load was 
24,000 lb. and after a load of 47,000 lb., no more load 
could be added, the bar finally breaking between the punch 
marks, when the load was 42,000 lb. The broken pieces 
were placed end to end, and the distance between the punch 
marks was found to be 10.4 in. Required the elastic limit, 
ultimate strength, and the ductility of the material. 



APPLIED MECHANICS 21 

24. If the bar in problem 1 has the load increased 
from 60,000 to 120,000 lb., how much more work is done ? 

25. How much should the bar in problem 24 stretch 
while the additional load is being added ? 

26. A certain grade of piano wire has an elastic limit 
of 100 tons / sq. in. If the diameter of a piece of the 
wire is .05 in. in diameter, required the diameter of a 
wrought iron wire to carry the same load when the unit 
stress is equal to the elastic limit in each case. 

27. Show that the work done by an axial force on 

1 S^ 
any bar is -Sr= - — x volume, provided the elastic limit is 

not passed. 

28. If the load in problem 1 is suddenly applied, 
what will be the value of the maximum stress induced 
in the bar? 

29. If a square wrought iron bar is to sustain a sud- 
denly applied load of 60,000 lb. and the stress is not to 
exceed 15,000 lb. / sq. in., required the dimensions of the 
bar. 

30. A round steel rod is to carry a tensile load of 
37,700 lb. with a factor of safety of five ; required the 
diameter of the bar. 

31. Find the factor of safety in problem 1. 

32. A steel rod 2 in. in diameter in a bridge truss 
has a unit stress due to the weight of the bridge of 
4000 lb. / sq. in. A heavily loaded truck, if placed on 
the bridge, will add 51,000 lb. load to that already on 
the rod. Is it safe for the truck to cross ? 



CHAPTER II 
APPLICATIONS 

Article 23. Bars of Uniform Strength. 

In the previous chapter, the bar was one of uniform 
section and no account was taken of its weight. When 
the bar was short, the effect of the weight of the bar 
could be neglected in comparison with the applied loads, 
and the unit stress found was that due to the loads alone. 
If the bar under axial forces is very long, the stress due 
to its own weight becomes too large to be neglected, and 
the stress in the bar is that due to the loads plus that due 
to its own weight. 

Take the case of a wire rope used to hoist a bucket from 
a deep mine shaft. The weight of the bucket and its 
contents produces a certain unit stress in the rope that is 
equal at all sections of the rope. 

If P is the weight of the bucket and its contents, and 

P 

A the area of the section of the rope, this stress is S = —-, 

The sectional area of the rope at any point has to support 
the weight of rope below that point, as well as the weight 
of the bucket and its contents; therefore the section of 
the rope at the upper end being A^ and IF the weight of 

the rope, S= — —^ — instead of — —. 
A A 

It is readily seen that if W is small, the value of JP + TF 

22 



APPLICATIONS 23 

is not sensibly greater tlian the value of P ; the value of S 

p 

will not be materially changed from — -. 

■^ 

In the case where P + TF is much greater than P, and 
the rope is the same size throughout, it must be large 
enough to carry a load of P -\-W, 

When a long vertical bar of uniform section is under an 
axial load, it follows that every section, except one, is 
larger than necessary, and if the section area is varied so 
that the unit stress in each section is the same, the weight 
of the bar could be reduced. Such a bar is called a Bar 
of Uniform Strength, This does not mean that the 
strength of the bar is the same at all sections, but that 
the change of section area makes the unit stress the same 
at all sections, and might better be termed a bar of uni- 
form stress, or a uniformly safe bar. 

Consider such a vertical bar, length ?, and an axial load 

P 

P, The smallest possible area, JLq, is given by ^ = -^» 

where S is the allowable unit stress, and design the bar 
so that S shall be constant. In Fig. 23, let A^ be the 
area at the end where the load is applied, and A be the 
area at any distance y from that section. Then at a dis- 
tance y •\- dy the area must be increased to ^ + dA. Let 
w equal the weight of a cubic unit of the material; 
then the additional weight to be carried on the area 
A + dA is Aw dy^ since the term containing dA dy 
can be neglected in comparison to the term containing 
only dy. Since 8 is constant, and this weight is to be 
carried on the area dA^ 

■.A Aw dy -J jS dA rt N 



24 



MECHANICS OF MATERIALS 



gives the relation between the increase of length and the 
increase of area. If (1) is integrated, 



w 



(2) 



S 



and since A = J.^, when «/ = , 0= log A^. 

Substituting this value for C in (2) and transposing, 
log,^ = -^+log,^oOrlogio^ = 0.434-y + logio^ (3) 
is an expression for the relation between the least area and 





Fig. 23. 

the area at any distance y from that section. In the 
application of the formula to a given case, different values 
might be assigned to ^, and the corresponding values of A 
found, enough values being calculated to enable the pro- 
file to be drawn. In this case tlie outline of a vertical 
section is slightly curved. If the vertical section is made 
trapezoidal, and the bar is a masonry pier, the top of 



APPLICATIONS 



25 



the trapezoid is made proportional to Aq^ and the base to 
the value of A in equation (3), when ?/ is the height 
of the pier. 

Such a pier would require more material than one of 
uniform strength ; but, although the unit stress would be 
only approximately equal at all sections, it would repre- 
sent common practice. 

Art. 24. Thin Pipes, Cylinders, and Spheres. 

Take a pipe, internal diameter _D, thickness of the pipe 
wall t, carrying a water pressure of H lb. / sq. in., to find 




Fig. 24 a. 

the unit stress in the walls of the pipe. As each unit of 
length of the pipe is under the same forces, we may take 
the length as unity. Suppose a length of pipe equal to 
unity (Fig. 24 a) to be cut by a diametral plane X-X; 
then the stresses acting on the pipe walls at the section 
cut by the diametral plane must resist the pressure of the 
water tending to force the two halves of the pipe apart, 
and if ^ is small, the stress may be considered as being 
uniformly distributed over the sections of the pipe walls 
cut by the diametral plane. Therefore, if we can calcu- 
late the value of P, the force tending to separate the two 
halves of the pipe, the formula P = AS^ where A is the 



26 



MECHANICS OF MATERIALS 



area of the section of the pipe walls, will give the required 
unit stress. 

A principle of hydraulics states that the pressure of 
water is the same in all directions and normal to the sur- 
face. Let Fig. 24 h represent a half section, perpendic- 
ular to the axis of the pipe. If 6 is any angle, then 






?^ 



:<s 



ft 




>RrdeCo36 



Fig. 24 6. 

for a length of pipe equal to unity, the radius of the pipe 
being r, an area of the internal surface of the pipe, equal 
to rd6^ carries a pressure of H lb. / sq. in., or the total 
radial force on that area is Rrdd. This force may be 
resolved into components,* parallel and perpendicular to 
the line X-X^ which is the trace of the diametral cutting 
plane. The components are Rr cos Odd and Rr sin 6d0. 
The sum of the Rr cos Odd forces for one half of the pipe 
is zero, and the sum of the Rr sin Odd forces for the same 
half is 2 Rr^ or RB^ which is the force per unit of length 
perpendicular to the cutting plane, resulting from the 
internal pressure R. 

Substituting this value in the general formula P = AS, 



APPLICATIONS 27 

and noting that the area of the section of the pipe walls is 
2 1, we have BI) = 2 St (1) 

as a general expression for the unit stress induced in a 
longitudinal section of a pipe whose walls are thin. If 
the ends of the pipe are closed, the internal pressure of 
the water on the ends of the pipe tends to rupture the 
pipe in a plane perpendicular to the axis. 

The force acting on the ends of the pipe is evidently 

, and the area to resist this force is irDt ; hence a 

4 

substitution of these values in P = AS gives HI) = 4 St, 
showing that the stress in a plane perpendicular to 
the axis is only one half that on a plane through the 
axis. 

For a sphere with thin walls, the water pressure tends 
to produce rupture on the line of a great circle. It is 
readily seen that the pressures and areas are the same 
as for a plane section perpendicular to the axis of a cylin- 
der ; hence the same relation holds true. 

Akt. 25. Thick Pipes. 

If t is large, the stress in a plane through the axis is 
no longer uniformly distributed over the area of the sec- 
tion, but is greater on the internal radius. 

Many formulas have been proposed for finding the 
maximum unit stress in this case, the one given here 
being due to Barlow. The results are in simple form, 
and the value of the maximum unit stress being greater 
than that given by the more exact discussions, places the 
error on the side of safety. 

Barlow's formula assumes that when the fluid pressure 
acts on the internal surface of the pipe, while the diameter 



28 



MECHANICS OF MATERIALS 



is increased, the volume of the pipe walls for a unit of 
length remains unchanged. 

If we let I) be the internal, and D^ be the external, 
diameters of a pipe whose walls are thick (Fig. 25), and 




Fig. 25. 



the thickness of the pipe walls ^, before the pressure is 
applied the volume of a ring ona unit in length is 



Let e and e-^ be the extensions of the diameters due to 
the fluid pressure and the volume becomes 



TT 



TT 



l(i>^ + e,y-'^(D + ey. 



(2) 



Expanding (2), neglecting the e^, as « is a very small 
quantity, and equating (1) and (2), the equation reduces 

*o D^e^^De. (3) 



APPLICATIONS 29 

The unit elongations are — and -J-, since the change in 

the circumference of the thin shells of diameters D and 
D^ are ire and tt^^ while the original circumferences are 
ttD and irD^ The unit stresses in the thin shells of the 
diameters D and i)^, being S and S^^ as the unit stresses 
are proportional to the unit deformations within the elas- 
tic limit, 

^ = ^=:^. (4) 

A 

But I>xH — -^^' hence, 

e, D' 



and substituting this value of — in (4) gives 

or the unit stresses are inversely proportional to the 
squares of the diameters or radii. 

Let S^ be the unit stress at a radius x. Then, from (5), 



and the total force exerted over the area dx times 1 is 

SJx^^Sr^^' (6) 

The integral of the left hand member of (6) is the sum- 
mation of all the stresses on one side of the pipe, and is 



30 MECHANICS OF MATERIALS 

equal to one half of the total fluid pressure tending io 
rupture the pipe ; hence the total force is 

DE=2Sr^r^='^r^^^^^. (7-5 

*^r x^ r + t D-\-2t ^ ^ 

Equation (7) reduces to ED^ = 2 St instead of EJ) = 2 St 
for thin pipes. Formula (7) is the one to use in all 
cases where the value of ED^ is enough larger than ED 
to cause serious error. The error in Barlow's formula 
increases as the internal radius decreases, and for thick 
pipes where the diameter is small, the more exact formulas 
of Lami and others should be used. (See Merriman's 
"Treatise on the Mechanics of Materials.") 

Art. 26. Riveted Joints. 

In the determination of simple stress, such as tension, 

P . 

compression and shear, the formula 'S=—r always gives 

the relation between the force P and the unit stress S. 
The area A must always be the area over which the stress 
is induced, and will, for tensile or compressive stresses, be 
a section of the bar perpendicular to the line of action of 
the force P, and parallel to the same line for shearing 
stresses. If the area of the section of the bar varies for 
different cutting planes, the plane that gives the least area 
should always be chosen. 

When two plates are joined together by means of rivets, 
the joint is called a Eiveted Joint. 

Art. 27. Tension in the Plates. 

Let A and B (Fig. 27 a) be two plates joined together 
by means of rivets passing through the cover plates a and b. 
Let P be the tensile force tending to separate the plates 



APPLICATIONS 



31 



>< 



B 



( 



a 



( 



) 



) 



V 




A and B, w the width 
of the plates, t the 
thickness of the plates 
A and B^ t^ the thick- 
ness of the plates a and 
5, and (^ the diameters 
of the rivets. 

Since P is a tensile 
force, and the greatest 
number of rivets in 
line is two, if we pass 
a plane perpendicular 
to the line of action of 
the force P through 
the line of the two 
rivets, cutting either 
of the plates A or B, 
or the cover plates a and b, the stress which acts in such a 
plane to resist separation is the product of the unit stress 
induced and the area cut. 

The area" cut by the plane is either the thickness of the 
two cover plates a and b times the width of the plates less 
the diameters of the rivets in line, or the thickness of the 
plates A or B into the same quantity. Hence the relation 
between the tensile force P and the unit stress induced 
in the plates is, 2ti(w — 2 d) iSf — P, or t(w ~2 d) S^ = P, 
depending on whether the failure is in the plates ^ or ^ 
or the covfer plates a and b. 

As there is no reason why one of these sections should 
be stronger than the other, 2 t^ is generally made equal to 
t. Since any other cutting plane would cut a larger area, 
the value of S as given in the above formula is the largest 



Fig. 27 a. 



32 



MECHANICS OF MATERIALS 



value possible with the force P. Figure 27 h shows the 
failure by tension in the plates. 



B 





rty=c^ 



V 



Fig. 27 6. 




Art. 28. Shear on the Rivets. 

If the plates do not yield in tension, in order to pull 
the plate A away from B^ and the cover plates, the num- 
ber of rivets passing through A must be sheared off in 
two sections parallel to the line of action of the force P, 
and perpendicular to the axis of the rivets. Therefore 
the area to resist shear on the rivets caused by the force 
P must be twice the sectional area of each rivet times the 
number of rivets passing through the plate A. These 
values substituted in the general formula, P = AS give 

(See Fig. 28.) 



for this joint, 2 x —^ — S^ 



p. 



APPLICATIONS 



33 



Art. 29. Compression on the Rivets or Plates. 

Suppose that the plates resisted the ten- 
sile stress, and that the rivets, the shearing 
stress, the force P acting on A^ causes the 
plates to bear on the cylindrical surface of 
the rivets. The exact effective area of each 
rivet, or the plate through which it passes, 
is not known ; but it is assumed to be the 
projected area of the rivet; that is, the 





diameter of the rivet 

times the thickness of 

the plate through 

which it passes. On 

this assumption, taking 

2t-^ = t^ the area to 

resist compression on 

each rivet is dt. This 

area times the number 

of rivets passing through A^ when 

substituted in the general formula, 

P = AS, gives for this joint, 

2 dtS, = P„ 

as the relation between the force P 
and the unit compressive stress on 
the rivets or plates. (See Fig. 29.) 
As there is no other way that the 
joint can fail, the equation that gives the least value of P 
determines the way in which the joint is most liable to fail. 

Art. 30. General Case of Riveted Joint. 

In general, while the joint may be very long, the rivets 
are regularly spaced. In this case, the distance between 



34 MECHANICS OF MATERIALS 

the centers of any two rivets in line is called the " pitcla " 
of the rivets, and P is taken as that proportion of the load 
on the entire joint that the pitch is of the length; or in 
other words, P is the load or force on the joint for a 
distance eqnal to the pitch of the rivets. 

Tension in the Plates, 

Taking P as above, and letting p be the pitch of the 
rivets, the relation between the tensile unit stress in 
the solid plate and the force P is, 

tp8,=^P, (1) 

where, if aS' is the safe unit stress, P is the safe load. 

For all joints in tension, since there can never be but 
one rivet in line in the distance p^ 

tip-d)8, = P,. (2) 

If, as before, aS'^ is the safe tensile unit stress, P^ is the safe 
load when failure is considered as taking place by tension 
in the punched plates. If the values of Sf are the same 
in (1) and (2), it is easily seen that P^ can never equal P, 

Shear on the J^ivets. ^ 

The relation between the load P and the unit shearing 
stresses will depend on the nature of the joint. In any 
given case, the product of the number of times each rivet 
niay shear, the number of rivets in the distance p, and the 
area of the section of the rivet perpendicular to its axis, 
will be the area over which the shearing stresses act. 
Letting e be the number of rivets times the number of 
sections in the distance p^ then 

^ecPS;=P.. (3) 



APPLICATIONS 35 

AVlien safety of the joint against failure by the shearing 
of the rivets is considered, if Sg is the safe unit shearing 
stress, then P^ is the safe load. 

Considering (2) and (3), the values of p and d can be 
chosen so that when safe values of the unit stresses S^ and 
Ss are used, P^ = P^, but they are not necessarily equal. 

Compression on the Rivets or Plates. 

Taking S^ as the safe unit stress in compression, and 
td times the number of rivets in the distance p as the area 
resisting compression, and letting c-^ be the number of 

'^i^e*''' c,tdS, = P, (4) 

is the relation between the unit stress in compression and 
the load P^. As before, P^ may have different values 
from either P^ or P^, but if they are assumed to be equal, 
and safe values of aS^, S^^ and S^ are used in equations (2), 
(3), (4), as there are three equations and three variable 
quantities, p^ t, and d can always be determined. 

If the values of p^ t, and d are found in this manner,, 
the joint will be equally safe against failure in all ways. 
In general, the equation which gives the least value of P 
will show the way in which failure is most liable to occur. 

Art. 31. Kinds of Riveted Joints. 

Lap Joints. Here the two plates to be joined together 
lap by each other and the rivets pass through both plates. 
The rivets tend to shear on but one section, and are said 
to be in "single shear." 

Putt Joints with Single Cover Plates. Here the plates 
are both in the same plane, and the joint between them is 
covered by a plate of the same thickness as the plates. 
Any rivet passes through the cover plate and one of the 



36 



MECHANICS OF MATERIALS 





Double riveted lap joint. 



Double riveted butt joint with 
single cover plate. 




I 



11 



1^ 



Triple riveted lap joint. 




Double riveted butt joint with double cover plates. 
FiQ. 31. 



APPLICATIONS 37 

plates that are to be joined together. The conditions for 
shear and compression are evidently the same as for lap 
joints. 

Butt Joints with Double Cover Plates. In this kind of 
a joint the plates are in the same plane, and the cover 
plates, each one half the thickness of the plates to be 
joined, are placed on either side of the joint. Any rivet 
passes through both cover plates and one of the plates that 
are to be joined. 

An inspection of the figure w^ill show that each rivet is 
liable to be sheared in two sections and is said to be in 
" double shear," while the conditions for compression are 
the same as for the one with single cover plates. 

Either type of a joint may have one or more rows of 
rivets, and the pitch in all rows is generally the same. 
The joint is said to be Single., Double., or triple riveted, as 
there are one, two, or three rows of rivets. The figures 
show the details of the various joints and styles of riveting. 
It is evident that if lines are drawn passing through any 
two adjacent rivets in the same row, and parallel to the 
line of action of P, the rivets included between these lines 
will be the number of rivets that are to be considered as 
resisting the shear and compression. 

Art. 32. Efficiency of a Riveted Joint. 

When P^, P^, and P^ are the maximum safe loads a 
riveted joint will carry, the efficiency of that joint may be 
defined as the ratio of the least of the above values, to the 
load the unpunched plate of the same length will carry 
under the same conditions. 

From this definition it is evident that the efficiency of 
any joint is P^, P^, or P^., divided by P, depending on the 



38 MECHANICS OF MATERIALS 

relative values of P^, P^,, and P^. Of all the ways in 
which a riveted joint may fail, the failure b3^ compression 
of the rivets or plates is the least understood, and many 
engineers design the joint for equal strength in tension 
and shear, and simply check the resulting dimensions for 
the compressive stress. This practice has resulted in the 

efficiency of a riveted joint being given as , but this 

P 
is only approximately true unless P^= P^=i P^. 

In general, when the values of t and d are calculated, 
the nearest commercial sizes have to be chosen, and the 
values of P^, P^, and P^ are rarely ever equal. 

In many cases the pitch is fixed by the conditions for 
the tightness of the joint against leakage, as for boilers, 
tanks, and pipes, and in such cases only two conditions 
can be satisfied. 

In the development of the preceding formula no account 
has been taken of the friction that must exist between the 
plates through which the rivets must pass. 

As there is no good theoretical way of introducing the 
resistances due to friction in the formulas for strength, 
riveted joints have been pulled apart in testing machines, 
and the accuracy of the formulas checked by the breaking 
load as determined by the test. While the results in 
many cases seem to show that the theory that has been 
given here does not hold true, the conditions that are con- 
current with the rupturing load not being the same as when 
all the stresses are within the elastic limit, there seems to 
be no good reason why the formulas as developed will not 
giye reliable results. 

In the design of a riveted joint for a pipe or a boiler to 
carry a pressure of R lb. /sq. in., as one half of the 



APPLICATIONS 39 

internal pressure tending to disrupt the pipe or boiler 
is carried on one joint of the shell the value of P to be 
used in the formulas is one half of the total pressure 

acting over a length jt?, or, — —^ = P. 

Art. 33. Stresses Due to Change of Temperature. 

All metals tend to change in length as their temperature 
changes. If the change is resisted, that resistance must 
cause a stress in the material. 

Consider a bar I units in length, A units in area, free 
to change its length as the temperature changes. If the 
change in length due to a given change of temperature is 
6, and a force is exerted to restore the bar to its original 
length, the unit stress induced by that force will be given 

e 
Therefore if a force prevents the change from taking 
place, it must induce an equal unit stress, and this unit 
stress is independent of the area of the bar. Knowing 
the change' in unit of length for a change of 1° of tem- 
perature, or the coefficient of linear expansion, the unit 
stress in any bar corresponding to any change of tem- 
perature may be found provided the unit stress is within 
the elastic limit of the material. 

If the bar is under an initial unit stress before the 
change of temperature, the change will increase or de- 
crease that stress, depending on the nature of the initial 
and temperature stresses. 



40 MECHANICS OF MATERIALS 

PROBLEMS 

1. How long will a bar of wood have to be, in order 

that its own weight will produce a unit stress of 300 lb./ 

sq. in. ? The bar is hanging vertically. 

Solution. Taking the weight of a bar of wood, 1 sq. in. in sec- 
tion and 3 ft. long, as |f lb., the bar will have to be as long as 300 
divided by if, equal 360 yd. Ans. 

2. What is the length of a vertical steel bar a square 
inch in area, that carries a tensile load of 4000 lb., at 
the lower end when the maximum unit stress is 15,000 
Ib./sq. in. ? 

3. Find the probable elongation in problem 1. 

Solution. Since the maximum unit stress is 300 lb. / sq. in. and 

the minimum 0, the average unit stress must be 150 lb. / sq. in., 

S I 
and we have given that E = — , 

e 

1,500,000 = 150 X 360 x 36^ .^ ^ ^ ^ 296 in. 

e 

4. Find the total elongation in problem 2. 

(Total elongation is that due to the load and its own weight.) 

5. Find the height of a brick chimney of uniform 
section, when the maximum compressive unit stress is 18 
tons /sq. ft. 

6. Suppose that the sectional area of the base of a 
chimney was twice that at the top, and that the change in 
area was uniform, how high could the chimney be built if 
the limiting value of the unit stress was 18 tons /sq.ft.? 

7. Find the areas of the top and bottom section of a 
stone pier, 100 ft. high, to carry a load of 240 tons, the 
unit stress in all sections to be constant. 

8. If the pier in problem 7 had the top and bottom 
areas as found and the vertical section was trapezoidal, 
find the unit stress at the bottom of the pier. 



APPLICATIONS 41 

9. The wire rope used for hoisting in a certain mine is 
1^ in. in diameter, and weighs 2.5 lb. /ft. If the mine 
is 800 ft. deep and the safe working load for the rope is 
5^ tons, what weight may be raised ? 

10. A pipe 6 in. in diameter is to carry water under a 
pressure of 1000 Ib./sq. in., with a factor of safety of 
6; required the thickness of the pipe walls. 

11. A standard 2-inch steel pipe is 2.375 in. outside 
diameter and 2.067 inside. This size is tested under a 
pressure of 500 lb. /sq. in.; required the unit stress in 
the pipe walls. 

12. A steel pipe 10 in. in diameter is to carry water 
under 2770 ft. head. The factor of safety is to be 10. 
Find the thickness of the pipe. 

(A column of water 1 ft. high and 1 sq. in. in area weighs .434 lb.) 

13. Check the results in problem 12 by Barlow's for- 
mula and find the unit stress. 

14. Compare the maximum unit stress in the pipe of 
problem 11, as determined by the formulas for thick and 
thin pipes. • 

15. Write the formulas for determining the strength of 
the following riveted joints intension, compression, and 
shear. The pitch is p, the thickness of the plates t, diam- 
eter of the rivets c?, and the safe unit stresses in tension, 
compression, and shear are aS'^, /S'^, and Sg. 

(a) Single riveted lap joint. 

(5) Double riveted lap joint. 

(c) Single riveted butt joint with one cover plate. 

(d) Double riveted butt joint with one cover plate. 

(e) Single riveted butt joint with two cover plates. 
(/) Double riveted butt joint with two cover plates. 
(^) Triple riveted butt joint with two cover plates. 



42 MECPIANICS OF MATERIALS 

Solution for (a). ^ 

If Pt is a tensile force acting on the joint for a distance equal to 
the pitch, then since P = AS, and the area to resist the tensile 
stress is t(p — d), Pt = t(p — d)St gives the relation between the 
load and the unit tensile stress. 

Let Pc be the tensile force that brings compression on the rivets. 
As there is but one rivet in the distance j9, the area to resist compres- 
sion is td, and since P = .4^', P^ = tdS^ is the relation between the 
load and the compressive unit stress, and if Pg is the tensile force that 
produces shear on the rivets, as there is only one rivet in the distance 

», and it can shear in but one section, as P = AS, P, = Ss is the 

relation betw^een the load and the unit stress in shear. 

16. Assume that the values of p, t, d, St, Sc, and >Ss for 
any given joint are taken so that Pt = Pc = Ps show that 

CiSc 

the efficiency is given by — - — ^—^ 

ciSc~]~ St 

Solution. Since Pt = Ps 

t{p-d)St = citdSc 

(ciSc-\- St)d 
P = ^ 

Since Pt, Pc and Ps are all equal, 

efficiency = . 

P 
Substituting for p, 

ciScD 
efficiencv = ^ ^' - ""'^^ 



(ciSci-St) ciSc-^St 



St ^ 
which is the efficiency for any joint for which the values 

of Pt, Ps and Pc are all equal. 

17. A steam boiler, 60 in. in diameter, carrying 120 
Ib./sq. in., is to have the longitudinal seams double riv- 
eted butt, joints with two cover plates. Take S^ = 12,000 
Ib./sq. in., aS'^ = 10,000 Ib./sq. in., and make the joint 
equally safe against failure by either tension or shear. If 



APPLICATIONS 43 

the efficiency is to be approximately 75%, required the 

thickness of the plates, the diameter, and pitch of the rivets. 

Solution. The thickness of the plate is given by RD = .75 • 2 St 
as the unit stress in the unpunched part of the plates can be only 
75% of the allowable unit stress, or, 

120 X 60 = .75 X 2 X 12,000 t 

120 X 60 _ Q ^ 
.75 X 2 X 12,000 ' ' 

and the expression for the efficiency in tension being efficiency 

P^^ = .75, 
P 

p = 4:d. 

Take y'g as the nearest market size for the required thickness of the 
plate, for equal strength, 

t(p-d)St = 2x2 x^Ss, 

tV (^d- rf) 12,000 = 7rd^ X 10,000, 

d = I'' approximately ; 
then t = ^\", d = i", and p = 2". 

18. Using the approximate values for p, t, and d as deter- 
mined in problem 17 and assuming aS = 10,000 Ib./sq. in., 
find the other allowable unit stresses, considering the joint 
to be equally safe against failure by tension, compression 
and shear. 

19. A boiler, 30 in. in diameter, has double riveted lap 
joints, plates ^ in. thick, rivets ^ in. diameter, pitch of 
the rivets 2.5 in. Taking S^ as 60,000 Ib./sq. in., find 
the pressure per square inch that may be carried with a 
factor of safety of 6, considering failure by tension of the 
plates alone. 

20. What are the values of S^ and jS^, and the efficiency 
of the joint, in problem 19? 

21. A triple riveted butt joint with two equal cover 
plates is to have an efficiency of 80 %. Using S^ = 10,000 



44 MECHANICS OF MATERIALS 

lb. /sq. in. as the safe unit stress in tension, what witi be 
the values of S^ and aS'^., when the joint is equally safe 
against failure bj tension, compression, or shear ? 

22. If Sf is taken as 10,000 Ib./sq. in., and S^ as 15,000 
lb. /sq. in., what is the highest possible efficiency of a 
double riveted lap joint, designed for equal strength 
against tension, compression, and shear? 

23. A steam boiler with double riveted lap joints is 
to carry 125 lb. /sq. in. pressure. The allowable tensile 
unit stress is 10,000 lb. /sq. in. The plates are | in. 
thick, rivets 1 in. diameter, and the pitch 3J in. What 
will be the largest possible diameter of the boiler ? 

24. Compare the values of S^, S^, and S^ in problem 
23. 

25. A 30-foot steel railroad rail undergoes a change of 
temperature of 100° . If the change in length is pre- 
vented, what unit stress will be set up in the rail ? 

Solution. The coefficient of linear expansion for steel is .0000065 
per degree ; hence the unit elongation for 100° is .00065 in., 
and since 

S = Ec, 

S = 30,000,000 X .00065 == 19,500 lb. / sq. in. 

26. For electric railway work, the steel rails are often 
welded together. Assuming that there is no change of 
length, what is the maximum range of temperature allow- 
able if the unit stress is not to be greater than the elastic 
limit? 

27. The walls of a building had bulged out, and to pull 
them into place, five steel rods each two sq. in. in area 
were passed through from one wall to the other. The 
temperature was then raised 100° and the nuts on the 
rods tightened, so that the load on each bolt was" 1000 
lb. When the rods are at the original temperature, what 
is the maximum pull they could exert on the walls ? 



APPLICATIONS 45 

28. At St. Louis, Mo., a battery of steam boilers was 
connected together by a pipe in which there was no pro- 
vision made for expansion. The temperature of the 
steam was about 360° F. and that of the room, 100". 
Assuming that there was no change in length, what was 
the maximum unit stress in the pipe due to the change of 
temperature ? 

(a) Material of the pipe steel ? 

(5) Material of the pipe cast iron ? 



CHAPTER III 



BEAMS 

Art. 34. Kinds of Beams. 

When a bar is placed in a horizontal position, and acted 
on by forces perpendicular to the axis of the bar, it is 
called a Beam. 



C 



3 



Cantilever Beam. 
Fig. 34 a. 




Cantilever Beam. 
Fig. 34 &. 



m 



wm. 




I 



^flj 



■ 



Hi 



fl; 



Simple Beam. 
Fig. 34 c. 



Continuous Beam. 
Fig. 34 d. 



I 



If the beam has two supports on which it merely rests, 
it is called a Simple Beam. 

A cantilever beam has only one support, which is at the 
middle, or, what is the same thing, has one end firmly 
fixed in the wall, leaving the other end free. 

When a beam has both ends firmly fixed in the walls, 
or one end fixed and the other merely supported, it is 
called a Fixed or Restrained Beam. 

A beam supported at more than two places is called a 
Continuous Beam, 

46 



BEAMS 47 

Art. 35. Reactions at the Supports. 

The reactions at the supports are the forces acting 
between the supporting walls and the beam, and so far as 
the beam is concerned, they may be treated as vertical 
forces acting upward. 

Art. 36. Uniform and Concentrated Loads. 

The loads on a beam are the weights that the beam 
carries, and since the attraction of gravity always acts 
downward, they may be represented as vertical forces. 
When the load is distributed uniformly over the entire 
length of the beam so that each element of the length 
of the beam carries the same load, the load is said to be a 
Uniform Load. 

When a load is carried on so small a portion of the length 
of the beam that the effect of the weight acting as it does 
over that small portion may be assumed to have the same 
effect as a single force acting at the center of the load, it is 
called a Concentrated Load. 

Since a simple or cantilever beam under any loads may 
be considered as a body acted on by forces which keep it 
at rest, the laws relating to the equilibrium of forces must 
be satisfied. 

In general, all the forces, loads and reactions, will be 
vertical, and the magnitude and position of the loads will 
be known, so that the magnitude of the reactions may be 
determined by applying the laws relating to the equilibrium 
of parallel forces. 

These laws are: 

The algebraic sum of all the forces equals zero, and 
the algebraic sum of the moments of all the forces about 
any point equals zero. 



48 MECHANICS OF MATERIALS 

These two equations are sufficient to determine £he 
reactions for simple and cantilever beams, as there are not 
more than two quantities to be determined. 

For all other beams another condition is derived by the 
use of the theory of the Elastic Curve. (See Chapter IV.) 

The length of a simple beam is the distance between 
the supporting walls and the distance the beam projects 
beyond the wall for a cantilever beam. 

While in any case the beam must rest on the supporting 
wall for a finite distance, the point of application of the 
single force that is to replace the resultant of the forces 
acting between the beam and the wall is taken at the edge 
of the wall beyond which the beam projects. 

Let Fig. 36 represent 

^ J ^ "p-A"-^ ^ simple beam, length Z, 

I^^ weight TT, carrying two 



^m 



m--i"-^ i 



^ concentrated loads P^ and 

^^ n£ 1*2 at distances 'p-^ and 'p^, 

^i^- ^' from the right reaction, 

and the values of the reactions i?i and H^ are required. 

The weight W may be considered as a uniform load, 
and for equilibrium, 

i?i + i^2-A-^2-TF=0. (1) 

Taking moments about a point in the line of action of 
7^2, and giving the moment a positive sign when it tends 
to produce clockwise rotation, 

Ii,l^B,^-P^p,-P,p,-^^(i. (2) 

The term containing H^ is zero, therefore It^ may be 
found, and by substituting for H^ in equation (1), i?2 ^^^.y 
also be determined. 



BEAMS 



49 



R^ may also be found by taking moments about a point in 
the line of action of i?i, and this value used as a check. 

If the weight of the beam, TFJ included both the weight 
of the beam and a uniform load, the equations would have 
been the same. 

Art. 37. Vertical Shear. 

Let Fig. 37 a represent a simple beam, loaded with both 
uniform and concentrated loads, and Fig. 37 5 a cantilever 
beam with the same loading. 

Suppose either beam to be cut by a plane X—X perpen- 
dicular to the axis of the beam, at any distance x from the 



ccmrrrr^ 



K 

Fig. 37 a. 



R. 



r 




Fig. 37 6. 



left end of the beam, and consider the end marked E. This 
end is acted on by known forces, since, when the loads are 
known, the reactions can be found. Relative to the end 
marked F^ these forces tend to produce translation either 
up or down, the magnitude of the resultant force being 
the algebraic sum of all the forces acting on the part E. 

Similarly, considering the part marked F^ the resultant 
of all the forces acting on F tends to produce translation 
relative to F^ and these two resultants being equivalent 
to all the forces acting on the beam, must be equal and 
opposite in sign, since the beam is in equilibrium. 

These two resultants are a pair of shearing forces, and 
either one is the force producing shear in the plane X—X^ 
and if we wish to call one of these forces the Vertical 



50 MECHANICS OF MATERIALS 

Shear at the section X-X, it will be necessary to define 
vertical shear so that the sign will be determined as 
well as the magnitude. 

The vertical shear at any section of a beam is defined 
as follows : 

The Vertical Shear for any section of a beam is the 
algebraic sum of all the forces acting on that portion of 
the beam, lying to the left of that section. 

When the resultant force acts upward, the vertical 
sliear is considered positive, and negative when it acts 
downward. 

Art. 38. Bending Moment. 

If we take the sum of the moments of all the forces 
that act on the right and also those that act to the left 
of the section X-X (Figs. 37 a and 37 ¥) about a point in 
that section, the resulting moments must be equal and 
opposite in sign, and as either is a measure of the tend- 
ency of rotation to take place about a point in the plane 
X-X^ they are the bending moments for that section. 

In order to fully determine the bending moment both 
as to sign and magnitude, it is defined as follows : 

The Bending Moment at any section of a beam is the 
algebraic sum of the moments of all the forces, acting on 
that portion of the beam lying to the left of the section, 
moments being taken about a point in that section.* 

It is considered positive when the moment tends to 
produce clockwise motion and negative for counterclock- 
wise motion. The " section of the beam " as used in the 
definition of both vertical shear and bending moment 

*■ A cantilever beam is always considered as being fixed at the right 
end, leaving the left end free. If the beam projects toward the right, 
look at it from the other side. 



BEAMS 51 

refers to any plane section perpendicular to the axis of 
the beam, and the " forces acting to the left of the sec- 
tion " includes both the loads and reactions whose points 
of application are on the left of the section. 

Art. 39. Resisting Shear. 

Since the part of the beam on the left of the section 
X-X (Figs. 37 a and 37 5) is acted on by external forces, 
and as a part of the whole beam it is in equilibrium, there 
must be internal forces acting in the section X-X^ which 
taken with the external forces acting to the left of the 
section, constitute a system of forces in equilibrium. 
Suppose these unknown forces to be resolved into their 
horizontal and vertical components. Then, since equi- 
librium exists, the algebraic sum of the vertical compo- 
nents of the internal forces must equal the sum of the 
vertical forces, and since the external forces have no 
horizontal components, the sum of the horizontal compo- 
nents of the internal forces must be zero, and also, the 
algebraic sum of the moments of the external forces must 
equal the sum of the moments of the internal forces, mo- 
ments being taken about a point in the section X-X. 

From the first condition, if we give the name of Resist- 
ing Shear to the sum of the vertical components of the 
internal forces, 

The Vertical Shear = the Resisting Shear., 

and assuming the shearing forces to be uniformly dis- 
tributed over the area of the section, then 

F= AS, (3) 

where A is the area of the section and S is the unit shear- 
ing stress, and F"the vertical shear for the section. 



52 MECHANICS OF MATERIALS 

Art. 40. Resisting Moment. 

From the second condition, since the horizontal com- 
ponents of the. internal forces must be either tensile or 
compressive forces, the sum of the tensile Forces = the 
sum of the compressive Forces. 

Giving the name of Resisting Moment to the moments 
of the internal forces about a point in the section, the 
third condition for equilibrium states that, 

The Bending Moment = the Resisting Moment. 

The relation between the bending moment of the exter- 
nal forces and the unit stresses in the section considered, 
cannot be found by the laws of mechanics alone, as the 
distribution of the internal forces is unknown. The in- 
formation necessary may be derived from the results of 
experimental observations on beams while under the 
action of bending forces. 

When a beam is under the action of bending forces, it 
is observed that along the concave surface of the beam 
the fibers * of the beam are shortened, while those on 
the convex surface are lengthened, and that along a 
certain plane section of the beam there is no change in 
length. 

We know that a compressive force shortens, and that a 
tensile force lengthens, any bar on which it acts, and that 
where there is no deformation there can be no force act- 
ing ; therefore the stress on the concave surface must be 
compression, and that on the convex surface, a tensile 

* The word "fiber" as used here may be defined as a bar of elemen- 
tary sectional area and a length equal to that of the beam, the whole 
beam being composed of a bundle of such fibers. It is not necessary 
that the beam should be of a fibrous material in order that this conception 
should be true. 



BEAMS 53 

stress, while at the certain plane, called the Neutral Sur- 
face, there is no stress of any kind. 

When the loads were such that there was no unit stress 
greater than the elastic limit, it was observed that the 
deformation of any fiber was proportional to its distance 
from the neutral surface. 

If we call the trace of this neutral plane on any plane 
section of the beam perpendicular to the axis, the Neutral 
Axis of the section, since there is no unit stress greater 
than the elastic limit, it is evident that the unit stress at 
any point in that section, and consequently the forces pro- 
ducing that unit stress, must vary directly as the distance 
from the neutral axis. This assumes that JE is constant, 

since U = — , and when the location of the axis is known, 
e 

the unit stress at any point may be found. 

Let Fig. 40 represent any cross section perpendicular to 
the axis of the beam, and the line X-X the neutral axis. 
From the experimental observations we know that the 
greatest unit stress must be at the 
greatest distance from the neutral / ^ n \ 

axis ; and letting e be the distance / i 1 V 

from the neutral axis to the fiber / \ 

most distant from that axis, A the 
area of the section, dA the area of Fig. 40. 

any fiber, ^ the distance of that fiber from the neutral axis, 
and S the unit stress at a distance e from the neutral axis, 
then since the force varies as the distance from this neu- 
tral axis, the force at any distance ?/, acting on the area of 

any elementary fiber dA is —yd A, and — ) ydA is the sum- 

c c^ 

mation of the horizontal forces acting on the whole section. 



54 MECHANICS OF MATERIALS 

From the conditions of the problem, this sum is equal 

to zero, and as neither S or c can be zero, I ^dA must be 
zero. 

If we assume the density to be constant, this is the con- 
dition where the axis of moments passes through the cen- 
ter of gravity of the section; hence the neutral axis 
passes through the center of gravity of the section. The 

V 
force on any fiber being — ?/c? J., the moment of this force 

c 

about the neutral axis is — y'^dA, and the sum of the mo- 

c 

ments of these forces about a point in the section is 
— I 'if'dA, which is the Resisting Moment by definition. 

The Cy'^dA is defined in mechanics as the moment of in- 
ertia of the section about a gravity axis, and is repre- 
sented by the symbol I. 

Therefore, since the Bending Moment = the Resisting 

Moment, aj 

M = —. (d) 

c 

In this formula ikTis the bending moment of the exter- 
nal forces that act on the left of any section, Zthe moment 
of inertia of the section about a gravity axis perpendicu- 
lai" to the direction of bending, c the greatest distance of 
any fiber from the neutral axis, while S is the maximum 
unit stress in that section. 

The formula expresses the relation between the bend- 
ing moment and the unit stress in the section, and if the 
maximum unit stress in a beam is desired, M must be the 
maximum bending moment for that beam under the given 
loads. 



BEAMS 56 

As an aid to memory, attention is called to the simi- 
larity between this expression and the one derived for 
axial stress. 

P and M are the external forces, S in each case is the 

unit stress induced; and as - depends on the shape and 

c 

area of the section for its value, it may be considered as 
replacing A in the formula for axial stress. 

Art. 41. Use of Formula. 

In the derivation of the formula for axial stress, there 

was no consideration taken of the intensity of the stress 

or of the nature of the material, the formula holding true 

for all unit stresses and materials. When the formula 

Mc 
S = —r was derived, certain conditions were specified. 

They are : 

(1) The material was to be elastic, and since we assumed 
that the forces were proportional to the deformations, the 
modulus of elasticity must also be constant. 

(2) In order for I y'^dA to be the sum of the moments 

of the differential areas about the gravity axis, the material 
of the beam must have a uniform density. 

(3) No unit stress to be greater than the elastic limit. 
If the material and loading of a beam does not satisfy 

these three conditions, the formula S = — - will not give 
the true unit stresses. 

The Modulus of Rupture is the value of S as derived 

from S = -— , when M is large enough to rupture the 

beam. Since the formula only holds true for unit stresses 
within the elastic limit, the value of the modulus of rup- 



56 MECHANICS OF MATERIALS 

ture as an engineering constant is at least doubtful. In 

testing cast iron bars in bending, the breaking load at the 

center, which is of course proportional to the modulus of 

rupture, is taken as a measure of the quality of the material. 

The results of such tests are useful for comparison only 

when the tests are made on bars of the same length and 

size. 

Mc 
The formula S — -— , expressing as it does the relation 

between the bending moment and the unit stress in a 
beam, is used in all calculations for the strength, safety, 
and design of beams. When sufficient data are given to 
fully determine the value of M, the value of either S or 

— may be found. 

^ I 

The value of - depends on the form and area of the sec- 
e 

tion, and is called the Section Modulus. 

Mc 
In the application of the formula S = — — to any given 

case, the question of units becomes one of great importance. 

It does not make any difference in the effect of M, 
whether it is expressed in inch pounds, or foot pounds; 
but as S is the unit stress and is usually given in pounds/ 
square inch, the value of M must be expressed in inch 
pounds and c and /in inches, in order to get correct results. 

The ton could just as well be used as the unit of weight 
and the foot as a unit of length, but such practice would 
require a special table of the constants of materials, the one 
given in the Appendix being based on the pound and inch. 

As there are many different sections that have the same 
section modulus, the designer is called on to choose a form 
of section best suited to the conditions that exist in the 
case in hand. 



BEAMS 



57 



The value of / for a rectangular section, breadth 5, and 
depth c?, about an axis through the center of gravity 

parallel to the side 5, is — -^, and ^ is -• 



Substituting these values in tlie gen- 

Mc 
eral formula S = -—- gives the value 




of iif as 



6 



I 



Therefore, when the section is rec- 
tangular we see that the value of M 
for any given value of S increases 
directly as h and as c?^, and any in- 
crease in the value of d increases the 
strength more than a proportional in- - 
crease in the value of 5, while the 
weight of the beam will be the same 

for either case. 

a 



-2:02— 



CD 







Fig. 41 a. 



A practical limit of the ratio of t is about 6. 

r Z. 61 



WT. 29.3 TO 34.6 LBS. 



ti. 



-244^ 



/if 



iO 



H-^^ 



From the known condition 
that the unit stress in any point 
in the section varies as the dis- 
tance from the neutral axis, it 
is evident that the form of sec- 
tion which presents the great- 
est area where the unit stresses 
are large and a minimum area 
where they are small will be 
the best from an economic 
Fig. 41 h. standpoint. 

The common steel I beam, so called from the form of the 
section, is an example of this distribution of area. There 



KfC 



1_ 



'^ 



-3-k- 



58 



MECHANICS OF MATERIALS 



_ T. 97 

WT. 9.3 LBS-; 





"W 



dk 




is always a maximum vertical shear to be resisted in all 
beams, and care should be taken that the area of the sec- 
tion chosen is large enough to resist 
the shearing forces. In most cases if 
the beam is safe against the bending 
forces, it will be safe against the 
shearing forces, but the 
unit shearing stress 
should always be inves- 
tigated before the final 
decision is made as to 
the size of a beam. 

The hollow box or cored sections in cast 
iron machine parts subjected to bending 
forces represent the best practice on account 
of the large value of I relative to the weight. 

Plate girders, made 
by riveting angle 
irons to a steel plate 
called the web, making a beam 
whose section resembles that of the 
common I beam, are in common use in structural steel 
construction. 



> -^ 

-J Ol 
Ol - 



.52'^' 



1-88 

Fig. 41 d 



Plate Girder. 
Fig. 41 e. 



Art. 42. Shear and Moment Diagrams. 

If a line which may be either straight, curved, or 
broken be drawn so that the ordinate to that line from 
any point of a straight line representing the length of the 
beam equals the vertical shear or bending moment for that 
section of the beam, the resulting diagram is called a Shear 
or Moment Diagram, depending on whether the vertical 
shears or the bending moments are used as ordinates.* 
* This line will be called the shear or moment line. 



BEAMS 59 

To draw either diagram, the shears or moments for 
each unit of length of the beam might be calculated and 
the results plotted to scale, using as ordinates the values 
of the moments or shears and as abscissa the distances of 
the sections from the left end of the beam. A line 
through the points so located would be the shear or 
moment line, as the case might be. 

This is a tedious process and may be shortened by a 
study of the effect of the different kinds of loads on the 
form of the shear or moment line. 

Art. 43. Shear Diagram. 

For a concentrated load, the difference between the 
shears for sections taken just to the right and left of the 
point of application of the load is the magnitude of that 
load, since in the former case the " negative forces acting 
to the left of the section " are increased by the magnitude 
of the load over those for a section taken just to the left of 
the load. Therefore, the shear line will always contain 
a line perpendicular to the length of the beam under each 
concentrated load. 

When only concentrated loads are considered, the shear 
line between any two concentrated loads will be a straight 
line parallel to the length of the beam, since there is no 
change in the forces acting to the left of any section 
taken between the two loads. 

For uniform loads of w pounds per linear unit, the 
shear line will be a straight line inclined toward the 
right, as the resultant force on the beam due to the uni- 
form load is decreased by wx^ where x is the distance of 
the section from the left end. 

If there are concentrated loads as well as uniform loads, 



60 



MECHANICS OF MATERIALS 



the shear line will be straight and vertical under the 
loads and inclined between the concentrated loads. 

To apply these principles, let Fig. 43 represent a 
simple beam, carrying two concentrated loads, P^ and P^^ 
and a uniform load of w pounds per linear unit. 

Remembering the definition of vertical shear, it is 
easily noted that the shear at the left end is equal to 
the left reaction R^ which is plotted as AO. The shear 
between the left end and P^ is R^ — wx^ where x is the 
distance of the section considered from the left end. 

For a section distant p-^ from the left end, taken just 
to the left of P-^, the shear is less than the shear to the 
left end by wp^.^ and the ordinate to the line AB at any 

point will be the vertical shear 
for that section. For a section 
just to the right of P-^, the 
equation of the shear has 
become R^ — wx — P^, as the 
two sections taken to the right 
and left of P^ are considered so 
close together that the uniform 
load has not increased. The 
shear line will therefore drop 
to (7 on a vertical line through 
the point of application of P^. 
Between P^ and P^ the shear will decrease at the same 
rate as between R^ and Pj, since the load increases di- 
rectly as the distance and CD will be parallel to AB. 
Then comes the drop due to P2, and EF is parallel to AB 
and CD. FO must be the value of the right reaction, 
since the sum of the vertical forces must be equal to zero. 
It is evident that the consideration of more concen- 




FiG. 43. 



BEAMS 61 

trated loads would simply extend the diagram in a similar 
manner, and also, if the magnitude of the uniform load per 
unit of length should change at any point, the inclination 
of the shear line would also change at the same point. 

Art. 44. Moment Diagrams. 

The general equation of the bending moment for any 
section of a beam may be written from its definition. 
For a section of a beam distant x from the left end, the 
moment of the left reaction R-^ about a point in that sec- 
tion is R^x^ and the moment of the uniform load on the 
left of the section about a point in that section is the 

X . tux 

arm - times wx. or -— -, hence, 

2 2 

n.jr^2 fthe sum of the moments of the loads"! 

M = R^x — - — j that act to the left of the section I 

"^ [about a pomt in that section. J 

If there are no concentrated loads, the term ^containing 
the sum of the moments, etc., is zero, and for a cantilever 
beam the term containing the reaction R^ is zero, since 
there is no left reaction. 

In some cases the supports of a beam are not at the ends, 
and in that case the moment of the left reaction would be 
R-^(x—thQ distance of the reaction from the left end). 
When there are uniform loads on the beam, the above 
equation shows that M varies with x^^ hence the moment 
line will be a curve. 

If there are no concentrated loads, the equation of the 

curve will be the same at all sections, being a parabola 

whose equation is n,^ -r. wx^ 

M— R.x 

^ 2 

When there are concentrated loads in connection with 
the uniform load, the form of the equation changes at each 



62 MECHANICS OF MATERIALS 

concentrated load, and the moment line, while still para- 
bolic in form, has a different equation between each pair 
of concentrated loads. 

If only concentrated loads are considered, the equation 
of the bending moment is 

{the sum of the moments of the loads acting] 
to the left of the section about a point in l 
that section. J 

If a^j is the distance from P^ of any section of a beam 
taken between any two concentrated loads P^ and P^, the 
moment of P^ about a point in that section is P-^Xy It is 
evident that the equation of the bending moment for any 
section will be changed by the addition of PiX-^ the instant 
the section is taken to the right of P^.^ and as at that point 
x-^ is very small there will be no abrupt change in the value 
of the bending moment as the section passes under Py As 
the value of il[f depends on the first power of a?, it is evident 
that the form of the equation is that of a straight line. 

Therefore, w^hen only concentrated loads are considered 
the moment line will consist of a series of straight lines 
whose inclination changes at each concentrated load. If 
the loads are all concentrated, the bending moments may 
be calculated for sections under the loads and plotted to 
scale. Joining the points so plotted by straight lines 
will accurately determine the form of the moment line. 
If uniform loads are to be considered, the moment line be- 
tween any two loads being a parabola, the bending 
moments for enough sections between any two loads must 
be found to enable the curve to be drawn (Fig. 43). 

The value of Mh\ the general equation for the bending 
moment is zero when x is zero, hence M\^ zero at the left 
end. The moments of all the forces about any point being 



BEAMS 63 

zero, M must also be zero at the right end. That this 
is true for simple beams is so apparent that no proof is 
needed. In the case of cantilever beams with the right 
end fixed in the wall, if we remember that such a beam is 
but one half of a beam supported at the middle and that 
the right end of such a beam is, strictly speaking, the 
middle, the truth of the statement is e\ddent. 

Therefore the moment diagram will be a closed figure 
in all cases. 

Art. 45. The Relation between the Vertical Shear and 
the Maximum bending Moment. 

Given a beam carrying both uniform and concentrated 

loads, the value of the bending moment at any section 

to the right of all the concentrated loads considered is : 

wx 
M = Rix — - — Pi{x — pi)—P2{x — p2) . . . -Pn{x-pn) 

where pi, P2 - - - Vn are the distances of the loads 
Pi, P2 ' . • Pn from the left end of the beam and x is the 
distance of the section considered from the same end. The 
value of x for which M is a maximum is the value that 

renders —— =0 OT Rix — wx — Pi — P2 . . . Pn = 0. 
dx 

The left hand member of the last question is the ex- 
pression for the vertical shear for any section of a beam, 
therefore the value of x which makes the vertical shear zero 
renders the bending moment a maximum. 

As the equation for i!i^ was a general one and will apply 
to all kinds of beams and loadings, the results are true 
for all cases. 

The section of a beam where the bending moment is 
maximum is called the Dangerous Section^ and the prob- 



64 MECHANICS OF MATERIALS 

lem of finding this section is simply one of finding where 
the vertical shear passes through zero. 

, Drawing the shear diagram, if the shear passes through 
zero under a concentrated load, the dangerous section 
is determined at once. When the shear becomes zero 
between two concentrated loads, the general equation for 
the vertical shear may be written for that part of the 
beam and equated to zero. 

The value of x which satisfies this equation determines 
the dangerous section. 

Having found the dangerous section, the bending 
moment may be calculated for that section, and when this 

value of Mi^ substituted in the formula i)[f= ~, the value 

c 

of aS' will be the maximum unit stress in the beam. 

Art. 46. Relative Strengths of Simple and Cantilever 

Beams. 

Let the uniform load on either kind of a beam be W 
and a single concentrated load at the middle of a simple 
beam or at the end of a cantilever beam also be W', then 
if a be some number whose value depends on the kind of 
a beam and the way in which it is loaded, the maximum 

bending moment for the beam may be expressed as 

This value substituted in the general formula M= — , 

c 

gives = — , which may be written W= — — . The 

a c cl 

strength of a beam may be defined as the weight it will 

carry with a given unit stress. From the above equation 

for TFit is evident that the weight a beam will carry with 

a given unit stress depends on the value of ot, hence the 



BEAMS 65 

relative strengths of simple and cantilever beams loaded 
as above are directly proportional to cc. If the expres- 
sions for the maximum bending moments in simple and 
cantilever beams loaded with W as above are written, an 
inspection of the results will show that for 

a cantilever beam loaded with TFat the end a = 1, 

a cantilever beam loaded uniformly with W a = 2, 

a simple beam loaded with TTat the middle a= 4, 

a simple beam loaded uniformly with W «= 8. 

Art. 47. Overhanging Beams. 

Beams that overhang the supports are called Overhang- 
ing Beams. The fact that the reactions do not have their 
points of application at the ends of the beam does not 
prevent the application of the laws of mechanics to the de- 
termination of the magnitude of two reactions. Consider 
a beam that overhangs one or both supports and loaded in 
any way. Taking moments about a point in the line of 
action of one of the reactions as i^g' ^^^^ moment of H^ — 
the moments of the loads on the left of M^ + the moment 
of the loads on the right of M^ = 0, and as the sum of all 
the forces is zero, M^ -\- M^ = the sum of all the loads. 

These two equations will suffice to fully determine the 
reactions It^ and H^ when the magnitude and position 
of the loads are known. 

The shear and moment diagrams can be drawn by the 
same principles that were applied to simple and cantilever 
beams. 

In general, the vertical shear will pass through zero 
at two or more points, giving more than one value of x 
for which the bending moment is a maximum. The value 
of the bending moment at each of these points must be 



66 MECHANICS OF MATERIALS 

calculated in order to find the greatest bending moment 
in the beam. 

The maximum moments may be either positive or 
negative, but when using the greatest value of M in the 

formula M = — , the substitution is to be made witliout 
c 

regard to sign. 

As the sign of the bending moment changes from posi- 
tive to negative, its value must pass through zero, and the 
position of the section of a beam for which the bending 
moment is zero is called an Inflection Point. The position 
of an inflection point may be approximately located by an 
inspection of the shear or moment diagrams and the general 
expression for the bending moment for that part of the 
beam written. 

Equating this expression to zero gives the position of 
the inflection point accurately. 

Akt. 48. Beams of Uniform Strength. 

When the maximum unit stress in all sections of a beam 

is constant, the beam is said to be one of Uniform Strength. 

The beams so far discussed have all had uniform sections, 

and the value of - was the same for all sections. 
c 

The bending moment iHi" varies for all sections, and if S 

is to be constant, - must vary with M, since M = For 

c c 

any beam loaded in any way the bending moment for a 
section at any distance from the left end may be expressed 
in terms of a variable distance x and this expression 

equated to — 
c 

Assigning different values to x, the corresponding values 



BEAMS 67 

of - may be determined and the section of the beam at that 

"" . I 

point chosen to satisfy the value of - as found above. 

Beams of uniform strength may have any form of section, 
but they are usually made either rectangular in section or 

an approximation to such a section. 

I hcP' 

For beams of rectangular section - equals — ;- and 

M = ' in which either h or c? may be variable. Ex- 

6 

pressing ilf^in terms of the variable distance x^ the law of 
the variation of h or d^ as the case may be, determines the 
shape of the beam. 

At any section of the beam where iH/ = 0, there is no 
moment to be resisted, and so far as bending is concerned 
the area of that section can be made zero. 

In addition to the unit stress due to bending, there is 
at all sections of the beam a shearing unit stress due to the 
vertical shear at that section. 

If aS' is the allowable unit stress in shear the area of the 

section where iHf = is given by ^ = — -, where V is the 
vertical shear at that section. 

Art. 49. Moving Loads. 

In many cases the position of the loads is not fixed, and 
as the loads may occupy various positions, the value of M 
for finding the greatest unit stress in a beam must be de- 
termined from the position of the loads which gives the 
greatest bending moment. 

When the loads on a beam may change their positions 
they are called moving or Live Loads to distinguish them 
from stationary or Dead Loads. 

Assume a beam, length Z, and Pj and P^ two unequal 
loads that pass over the beam. See Fig. 49, 



68 MECHANICS OF MATERIALS 

Let z be the distance of the greater load P^ from the left 
end of the beam, and p be the distance between the two 
Q loads. The dangerous sec- 

-P— $ * tion Avill always occur under 

one of the loads, and assum- 



. 5 



Sk ■ J ing that it occurs under P-^ 

^^^- ^^- Let Q be the magnitude of 

the resultant of P^ and P^^ and x be the distance of its line 
of action from P^ ; then 

P^x =^ P^{p - x'), 

Pi + A Q 

Therefore the resultant of P^ and Pg acts at a distance 

P V 

z -\ — -^ from the left end of the beam. 

Taking moments about a point in the line of action of 
the reaction P^^ 

and B^z^Qz-^- -^P-z = M. 

M will be a maximum when 



dM 



===(?_2^^__^ = 0, 



dz II 

I P^p 

Therefore when ill/" is a maximum the middle of the beam 
is halfway between the resultant of tlie loads and the 
dangerous section of the beam. 



BEAMS 69 

It is evident from the form of the equations that the 
same results would have been obtained had there been 
any number of forces. 

The results were obtained on the assumption that the 
dangerous section occurred under the first load, and if 
this assumption is true, the vertical shear must change 
sign as the section is taken to the right or left of the 
load Pj. 

When there are but two loads, the dangerous section al- 
ways occurs under the left hand load when the two loads are 
equal, and under the heavier load when they are not equal. 

When there are more than two loads, the position ^of 
the loads that gives the greatest bending moment does 
not always have the dangerous section under the maxi- 
mum load, but the general law holds true that 

When the middle of the beam is halfway between the re- 
sultant of the loads and the dangerous section, a maximum 
bending moment occurs. 

Art. 50. Use of Formula. 

To find the position of a system of loads that causes 
the greatest bending moment, assume the loads to be 
so placed that when the dangerous section is assumed to 
occur under any load, the above criterion is satisfied and 
the vertical shear passes through zero, and calculate the 
bending moment for that position. The result will be 
the maximum bending moment that can occur and have 
the dangerous section under the load as assumed. 

Assuming the dangerous section to occur under any 
other load, the maximum moment for that position may 
be found. 

The position that gives the greatest value of M will be 



70 MECHANICS OF MATERIALS 

the one where the greatest bending moment occurs as 
the loads move over the beam. 

When a uniform live load moves over a beam, the 
greatest bending moment occurs when the load extends 
over the entire length of the beam. When the load only 
partially covers the beam, the maximum moment occurs 
at the dangerous section and the above criterion holds true. 

The greatest vertical shear caused by any moving load 
is found at the supports when the resultant of the loads 
is nearest to that support. 

Art. 51. examination 

1. When is a bar called a beam? A simple beam? a 
cantilever beam ? a continuous beam ? 

2. What is meant by, " the reactions at the supports " ? 

3. Define the term Uniform Load; Concentrated Load. 

4. Name the laws of mechanics that are used to deter- 
mine the reactions for a simple beam. 

5. Define Vertical Shear and Bending Moment. 

6. As, "the sum of the forces," as well as, "the sum of 
the moments of the forces," acting on each side of the sec- 
tion are equal, why is it necessary to use the expression, 
" acting to the left of the section," in giving the above 
definitions ? 

7. From the definitions of the bending moment and 
vertical shear, write the expression for each in terms of a 
variable distance from the left end of the beam. 

8. Why is it necessary to say, " moments being taken 
about a point in that section," in defining the bending 
moment at any section ? 

9. Define Resisting Shear; Resisting Moment. 



BEAMS 71 

10. Certain laws are deduced from experimental obser- 
vations made on beams under the action of bending forces. 
W hat are they ? 

11. What is the Neutral Surface of a beam? the Neutral 
Axis of a section of a beam ? 

12. Show that the neutral axis passes through the center 
of gravity of the section. 

13. Prove that S=^^, 

14. State clearly what each symbol in the equation 
S = means, and the units that should be used in sub- 
stituting for each. 

15. What conditions as to loads and material must any 

Mc 
beam satisfy in order that >S'=— r-will be true for that 

beam ? 

16. What is a Modulus of Rupture ? 

17. Define the term Strength of a beam. 

18. Show that the strength of any beam depends on the 
value of c£, where a is a number depending on the kind of 
a beam and the nature of tlie loading. 

19. What is a Shear diagram ? a Moment diagram ? 

20. Define the Shear and Moment line. 

21. The shear line for a beam carrying only concentrated 
loads consists of horizontal and vertical straight lines. Why? 

22. When only uniform loads are considered, the shear 
line is a straight line from end to end. Why ? 

23. Show that for a beam with both concentrated and 
uniform loads, the moment line will be a series of curved 
lines. If there are no uniform loads, show that the 
moment line will consist of a succession of straight lines 
at different inclinations. 



72 MECHANICS OF MATERIALS 

24. If any part of the moment line is straight and 
parallel to the line representing the beam, what can you 
say of the bending moments for any sections taken in that 
part of the beam ? 

25. Define the expression, 'Hhe Dangerous Section of a 
beam." 

26. Give the relation that exists between the maximum 
bending moment for any beam, and the vertical shear for 
that section. 

27. Show that the problem of finding the section of a 
beam where the bending moment is a maximum, is the 
same as that of finding the section where the vertical 
shear passes through zero. 

28. What are overhanging beams? 

29. How do they differ from simple beams? 

30. Show why the reactions may be found in the same 
manner as for simple beams. 

31. What is meant by the term Inflection Point ? 

32. If the vertical shear is zero at more than one sec- 
tion of a beam, how can you find the greatest bending 
moment for that beam ? 

33. When is a beam said to be one of Uniform Strength? 

34. Show that if the section of any beam is varied so 

that — varies with M^ the beam will be one of uniform 
e 

strength. 

35. When are the loads on a beam called Moving Loads? 

36. Give the criterion for the position of a system of 
moving loads that causes a maximum bending moment. 

, 37. If there is more than the one position of the loads 
that satisfies the criterion, how can you tell which position 
causes the largest bending moment? 



BEAMS 73 



PROBLEMS 



1. Find the reactions for a simple beam carrying a 
uniform load of w lb. /in. The length is I in. and the 
whole weight TF lb. 

Solution. As the whole load is W, the sum of the two reactions 
must equal W or B A- Ti — W 

Taking moments about a point in the line of action of itg, since the 
resultant of the uniform loads act at the center of the beam, 

^ 2 

R.= — = — and R^ = — . 

^22 ^2 

2. Find the reactions for a simple beam, length ?, carry- 
ing a single concentrated load P at a distance p from the 
right support. 

Solution. The sum of the reactions equals the loads ; hence 

i?i + 7^2 = P. 

Taking moments about a point in the line of action of jRg, 

R^l - Pp = 0, 



Substituting the value of R^, 
Pp 



R -^P 



+ R2 = P or R^ = p(l-P\ 

3. Find the reactions for a simple beam 10 ft. long 
carrying a uniform load of 500 lb. /ft. 

4. Find the reactions for a simple beam 30 ft. long 
carrying a load of 10 tons at a distance of 20 ft. from the 
left end of the beam. 

5. Find the reactions for a simple beam 30 ft. long 
carrying two equal loads of 5 tons each at 10 ft. from 
either end, and a uniform load of 500 lb. / ft. 



74 MECHANICS OF MATERIALS 

6. Find the vertical shears at the middle and at the 
ends of the beam in problem 3. 

Solution. From the definition of vertical shear, F at a section 
taken just to the right of R^ is the value of that reaction, or 2500 lb. 

At the middle of the beam the force acting upward is simply the 
left reaction, and the forces on the left of the section that act down 
are the unit loads on that part of the beam, or one half the total uni- 
form load. Therefore the vertical shear at the middle is 

2500 - 2500 = 0. 

At the right end just to the left of the right reaction 

2500 - 5000 = - 2500 lb. 

or the vertical shear at R2 is — 2500 lb. 

7. Find the vertical shears in the beam given in prob- 
lem 4, at sections taken just to the right and left of the 
load and at each end of the beam. 

8. Find the vertical shears in the beam in problem 5, 
at each end of the beam and at sections taken just to the 
right and left of each load. 

9. Find the bending moment at the middle of a simple 
beam, length I in., 

(a) for a load P at the middle. 

(5) for a uniform load of w lb. /in. 

P 
Solution for (a). The reactions are each—, hence by the defi- 
nition of the bending moment, for a section - from the left end of 

P I PI 
the beam M = — x - = — , as there are no other forces acting between 

R, and P. 2 2 4 

10. Find the bending moments at section just to the 
right and left of the load on the beam given in problem 2. 

11. Find the bending moment at the wall for a canti- 
lever beam, length I in., when the beam carries, 

(«) a uniform load of w lb. / in. 

(5) a single load P at the free end. 

(c) a single load P at ^ in. from the free end. 



BEAMS 



75 



Solution for (a). As there is no left reaction, the bending 
moment is the moment of the forces acting on the beam about a point 
in a section taken at the wall. The moments of all the uniform 
forces being equal to the moment of their resultant, let the resultant 
of the forces wl equal W ; then 

12. Find the bending moment at each load and at the 
middle of the beam given in problem 5. 

13. A cast iron bar 1 by 1 in. in section and 36 in. long 
is broken as a beam. The modulus of rupture is 35,000 
Ib./sq. in. Required the maximum bending moment. 

14. Draw the shear dia- 
grams approximately to scale 
for simple beams 30 ft. in 
length loaded with 

(a) a uniform load of 100 
lb. /ft. 

(5) a concentrated load of 
3000 lb. at the middle. 

((?) a uniform load of 100 
lb. /ft. and a load of 3000 lb. at the middle. 

(c?) two equal concentrated loads at 5 and 10 ft. from 
the left end. 



3000 

.0CO00C)QQ000000. 



tsoo]^ 





I 

B 



00 



Problem 14 a. 



B 



Solution for (a), 
from the left end is 



The vertical shear at any section distant x 
V = 1500 - 100 X. 



This is the equation of a straight line, and as V is 1500 lb. at the 
left support and — 1500 lb. at the right support, if AB is 30 ft., AC 
= 1500 lb., and BD = — 1500 lb., then the line through the points 
C and D will be the shear line, and the diagram ABDC will be the 
shear diagram. 

15. Draw the shear diagram approximately to scale for 
a cantilever beam 10 ft. long loaded with 
{a) a uniform load of 500 lb. /ft. 
(J)) a concentrated load of 500 lb. at the left end. 



76 



MECHANICS OF MATERIALS 



so 


o 




^ 




K 








s 

-500 
C 


jfi' ' 
*500 







(c) a uniform load of 500 
lb. /ft. and a concentrated load 
of 5000 lb. at the free end. 

(c?) two equal loads 2500 lb. 
each at 5 and 8 ft. from the free 
end of the beam. 



Problem 15 6. 



Solution for (b). The equation 
of the shear line is F= — 500, since there is no left reaction and the 
only load is 500 lb. at the end. 

Therefore, the shear line will be a straight line parallel to AB and 
at a distance — 500 from that line. A BCD is the shear diagram. 

16. Draw the moment diagrams approximately to scale 
for the beams as given in problems 14 and 15. 

Solution for the beam in 14 a. The value of the bending 
moment at any section x from the left end is 



M = R^x 



wx^ 



= 1500 a; - 100 — 



2' 



where x is in feet. Giving any values to x, the corresponding value of 
M may be found. 



When a: = 0, 

X = ^ ft., 
a; = 10 ft., 
a: = 15 ft.. 



M = 6250 ft. lb. 
M = 10,000 ft. lb. 
M = 11,250 ft. lb. 



3000 

nnnnnnnnnnno 



1500 



As the loads are symmetrical with the middle of the beam, the 
values of M for x equal 20, 25, and 
30 ft. will be the same as for x 
equal 10, 5, and ft. Let AB = 
80 ft. and CD, EF, GH, etc., repre- 
sent on some scale the values of M 
corresponding to x equal 5, 10, 15 
. ., . 30 ft. ; then a smooth curve 
passing through ADF, etc., will be 
the moment line, and the figure 
AHB will be the moment diagram. Problem 16. 




BEAMS 77 

17, Write the expressions for the value of M and V 
for any section of the beam, and determine the maximum 
bending moments and vertical shears, 

(«) for a simple beam loaded uniformly with W lb. 

(S) for a simple beam loaded at the middle with W lb. 

(tf) for a cantilever beam loaded uniformly with W lb. 

(c?) for a cantilever beam loaded with W at the end. 

The length of all the beams being I ft. 

Solution fok, (a). The expressions for M and Fmay be obtained 
from the general formula by making " the concentrated loads to the 
left of the section " equal zero. 



Therefore, M = R^x - 


wx"^ 
2 » 


and 


F = -Rj — WXy 




P wl 


Hence 


T7- wl 

V = — — wx. 



This expression is zero when x = -, and as ikf is a maximum when 

V is zero, substituting the value of x, which renders F= in the 
expression for M, gives 

T^ _ wlx _ wx^ _ wZ^ _ wP _ wl"^ _ Wl 

Evidently F is a maximum when x is zero. 

18. A simple beam is 20 ft. long and carries two con- 
centrated loads, one 100 lb. at 5 ft. and the other 500 lb. 
at 8 ft. from the left end, and a uniform load of 100 lb. / ft. 
extending over the beam for a distance of 12 ft. from the 
right end of the beam. Draw the shear and moment 
diagrams and calculate the maximum bending moment 
and vertical shear. 

19. A simple wooden beam 20 ft. long, 8 in. wide, 10 
in. deep, carries a uniform load of 80 lb. /ft. Required 
the maximum unit stress in the beam. 



78 MECHANICS OF MATERIALS 

Solution. For a uniformly loaded simple beam the maximum 

value of M is : 

Wl 80 X 20 X 20 X 12 .^ ^^^ . ,, 

= — = 48,000 m. lb., 

8 8 ' ' 



and 



d 
c 2 6 6 6 



I M^ bd'^ Sx 102 800 
12 

e Mc 48,000 X 6 oftm. / 

6 = = — = odO lb. / sq. in. 

/ 800 ' ■ 

20. A simple wooden beam, rectangular in section, and 
20 ft. long, is to be designed to carry a load of 240 lb. at 
the middle with a maximum unit stress of 300 lb. /sq. in. 

(d may be assumed to be equal to 6 h.) 

Solution. The maximum bending moment is : 

,^ PI 240 X 20 X 12 , / bd'^ d^ 
M = — ■ = and - = — = — , 

4 4 c 6 36 

, M I Mx3Q .^ 
and - — = -. .-. = d^. 

S c S 

Whence d^ = 240 x 20 x 12 x 36 ^ -^^^S, or d = 12, b = 2. 
4 X 300 ' 

21. A simple wooden beam is 1 foot square and 10 yd. 
long. What uniform load can it carry if the unit stress 
is not to exceed 300 lb. / sq. in. ? 

22. If the beam in problem 21 also carried a load of 
1000 lb. at the middle, what unifoim load may also be 
C9;rried if the unit stress is not to exceed 400 lb. / sq. in.? 

23. Two planks, 12 in. wide and 2 in. thick, are placed 
one on top of the other and used as a simple beam. They 
support a uniform load of 1000 lb. What is the maximum 
unit stress in the material? Assume that the surfaces in 
contact are frictionless and that I =30 ft. 

24. A simple beam of wrought iron is 4 in. wide, 6 in. 
deep, 12 ft. long, and carries a uniform load of 32,000 lb. 
Is it safe? 



BEAMS 79 

25. If the beam in problem 24 was a cantilever beam, 
what uniform load will it carry with a maximum unit 
stress not greater than the elastic limit ? 

26. A simple steel 10 in. I beam weighing 25 lb. / ft. 
is 18 ft. long and carries a uniform load, including its own 
weight of 15,000 lb. Required the maximum unit stress. 

Solution. The table gives the value of - for this beam as 24.4. 

The maximum value of M is ^ 

Wl 15,000 X 18 X 12 , e ^^ 
— =: ■ — and S = . 

8 8 / 

TT o 15,000 X 18 X 12 X 1 -,a «AA lu / 

Hence S = — ^ = 16,600 lb. / sq. m. 

8 X 24.4 ^ ^ 

27. Show that when the weight TF of a simple beam is 
2 % of the load at the center, the error in the unit stress 
as found by neglecting the uniform load due to the weight 
of the beam is about 1 %. 

28. A common rule states that when the load at the 
center of a simple beam is greater than five times the 
weight of the beam the weight may be neglected when 
making the calculations for strength. What maximum 
error will this rule allow ? 

29. If W is the w^eight of a cantilever beam and -P the 

P 
load at the end, find the ratio of -^ when the error in 

the unit stress caused by the neglect of W is 5 %. 

30. Select a simple I beam of structural steel 24 ft. 
long to carry a load of 12,500 lb. at the middle with a 
factor of safety of 4. 

The maximum value of M is 

M = ^' = IMOO ^ 24 X 12. s = 6W0 ^ ^, ,b / _ .^ 
4 4 4 

M^I^ 12,500 X 24 X 12 ^ ^^ 

S c 4 X 15,000 

The table gives the value of / for a 15 in. beam weighing 45 lb. /ft. 

as 60.8, and as P = 15 W, this beam will satisfy the conditions. 



80 MECHANICS OF iAIATERIALS 

31. Select a standard steel I beam 10 ft. long to be used 
as a cantilever beam, to carry a load of 4 tons at the free 
end. Factor of safety 4. 

32. Select a simple steel I beam for a span of 20 ft. 
to carry two equal loads of 2 tons each at 5 ft. from either 
end and a uniform load of 300 lb. /ft., the unit stress 
not to be greater than 16,000 lb. /sq. in. 

33. Select a standard steel I-beam to be used as a simple 
beam 16 ft. long and carry a uniform load of 10,000 lb. 
Safe unit stress equal 16,000 Ib./sq. in. 

34. A simple steel I-beam 10 in. deep, 30 ft. long, weigh- 
ing 40 lb. /ft. supports a uniform load of 11,250 lb. If 
the allowable unit stress is 16,000 Ib./sq. in., is the beam 
safe? 

35. A cantilever beam 20 ft. long carries a uniform load 
of 50001b. Assume the unit stress equal to 16,000 Ib./sq. 
in. and select a steel I-beam to carry the load. 

36. Select a standard steel channel 12 ft. long to be 
placed with the flanges vertical and used as a simple beam 
to carry a uniform load of 15,000 lb. Factor of safety 4. 

37. A beam 30 ft. long is supported at points 10 and 5 
ft. from the right and left ends. There is a uniform load 
of 500 lb. / ft. between the supports and concentrated 
loads of 450 lb. at either end of the beam. 

(a) Draw the shear and moment diagrams. 

(^) Find the greatest bending moments and vertical 
shears. 

(c) Find the inflection points. 

(c?) Select a steel I beam to carry the loads with a 
maximum unit stress of 16,000 lb. /sq. in. 

38. A beam supported at two points 18. ft. apart over- 
hangs each support 6 ft. The overhanging ends carry a 



BEAMS 81 

uniform load of 300 lb. / ft. and there is a concentrated 
load of 12,000 lb. at the middle of the beam. 

(a) Draw the shear and moment diagrams. 

(5) Find the greatest bending moment and vertical 
shear. 

(c?) Find the inflection points. 

(d) Select a steel I beam to carry the loads with a 
maximum unit stress of 16,000 lb. /sq. in. 

39. Three men carry a stick of timber 12 x 12 in. x 12 
ft. long. One man is at one end and the other two are at 
such a point that each of the three men carries an equal 
load. Find that point. 

40. A cantilever beam of uniform strength rectangular 
in section is 12 ft. long and carries a load of 1200 lb. at 
the free end. The material is cast iron and the factor of 
safety is 10. 

Find the largest and smallest sections and make sketch 
showing the plan and elevation of the beam when, 
(a) the width is constant at 4 in. 
(5) the depth is constant at 12 in. 

41. A simple beam of uniform strength is rectangular 
in section and 12 ft. long and carries a uniform load of 
9600 lb. The material is cast iron and the factor of safety 
is 10. Find the smallest and largest sections and make a 
sketch showing the plan and elevation of the beam when, 

(a) the width is constant at 4 in. 
(5) the depth is constant at 12 in. 

42. If the beam in problem 38 was a rectangular steel 
beam of uniform strength and constant depth, find the 
proper size for the largest section when h = d for that sec- 
tion and the safe working unit stress is 16,000 lb. / sq. in. 
If the allowable unit stress in shear is 10,000 lb. / sq. in., 
find the area of the least section possible. Make a sketch 
of the plan of the beam. 



82 MECHANICS OF MATERIALS - 

43. Two equal loads are 6 ft. apart. Find their position 
as they are moved over a simple beam 20 ft. long that 
gives the greatest bending moment in the beam. 

44. Three loads each 4 ft. apart are moved over a beam 
18 ft. long. From left to right the loads are 4000, 2000, 
and 2000 lb. Find the position of the loads that gives 
the greatest bending moment in the beam. 

45. Two loads 6000 and 4000 lb., 5 ft. apart are 
moved over a simple beam 32 ft. long. What size steel 
I-beam will be required? The unit stress is 16,000 Ib./sq.in. 
and the stress due to the weight of the beam is neglected. 

46. As the loads given in problem 45 pass over the 
beam, asssume that the weight of the beam is 1600 lb. 
and draw a diagram to scale, using as ordinates the bending 
moment under the 6000 lb. load and as abscissa the dis- 
tance of that load from the left end of the beam. If the 
unit stress is 16,000 Ib./sq. in., determine the si^e of the 
steel I-beam required. 

47. If Q is the resultant of a system of loads moving 

over a simple beam, x the distance of Q from the left end, 

and a the distance of Q from the middle of the beam, 

show that the conditions for maximum moment require 

that I 

I — X = X — 2 a when a: > -, 

2 

I — X = X -}- 2a when x<-^l> 

and if this condition is satisfied, that the value of the 
maximum moment is given by 

wr ^ ( 9 ^2 I ^^® moments of the loads on 1 

-^'^max == y V^ ± ^ ^^ "■ 1 the left of the dangerous section.] 



CHAPTER IV 



TORSION 



Article 52. Derivation of Formula. 

In the previous chapters the forces were assumed to 
act in a plane passing through the axis and were either 
parallel or perpendicular to the axis. 

The forces that produce the stress known as torsion act 
in planes that are perpendicular to the axis of the bar, 
and while the lines of action of the forces are perpen- 
dicular to it, they do not pass through the axis. 

The effect of such forces must be to twist the bar. 
Assume a cylindrical bar, one end of which is firmly 
fixed in the wall, to be 
acted on by a couple ly- 
ing in a plane perpendic- 
ular to the axis of the bar 
at a distance I from the 
wall, and whose moment 
about that axis is Pp. 
A fiber of the bar that 
before the application of the force occupied the position 
of the line ad (Fig. 52 a), after the force has been applied 
will occupy the position of the helix ah^ and a point d on 
the surface of the bar will have moved to the position h. 

It is evident that the angle had^ the angle of the helix, 
is independent of the length of the bar and depends only 

83 




Fig. 52 a. 



84 



MECHANICS OF MATERIALS 



on the twisting forces and the material of the bar. Since 
any plane section perpendicular to the axis of the bar 
between the wall and the couple would contain an arc 
similar to hd^ and the length of that arc is proportional 
to the distance from the wall, the angle hod is propor- 
tional to the length of the bar and the twisting forces. 

Tills angle is called the angle of twist, and will be 
denoted by 6, 

By analogy with tension, the distance a point on the 
end of the bar moves under the action of the twisting 
forces being similar to the distance a point on the end 
of a tension bar moves under the action of the tensile 
forces, the ai'c hd may be taken as a measure of the 
deformation of the surface fibers of the bar due to the 
twisting forces. 

As the arc hd was proportional to the length, — can 
be taken as representing the unit deformation. 

Experiment has proven that when a bar is circular in 
section and no stress is greater than the elastic limit, the 

line od. moved to anv 
new position during the 
twisting of the bar, re- 
mains a straight line. 
This being true, the 
length of any arc d^-^ 
(Fig. 52 5), with a center 
at and a radius «/, is 
proportional to its radius, 
and as the arc is propor- 
tional to the deformation 
at the radius «/, the force 
Fig. 52&. producing this deforma- 




TORSION 85 

tion is proportional to the same distance. This reasoning 
is true for any perpendicular section of the bar between 
the plane of the couple and the wall. 

Assume the bar to be cut by a plane perpendicular to 
the axis between the wall and the twisting couple, and 
introduce forces in that section to render the free end in 
equilibrium. These forces must all act in a plane per- 
pendicular to the axis, since the external couple has no 
component perpendicular to such a plane, and their re- 
sultant must be equivalent to a couple whose moment is 
equal to that of the twisting moment. 

Considering the two faces of the bar in any section, 
it is evident that the face on the right end tends to slide 
against the face on the left end as the former tends to 
rotate about the axis of the bar, thus producing a shearing 
stress throughout the section. Since the force, and con- 
sequently the unit shearing stress, is proportional to the 
distance from the center of the bar, if we let S^ be the 
unit shearing stress in the surface fibers, e be the distance 
of those fibers from the axis of the bar, and ?/ be the dis- 
tance of any fiber from the same axis, then — ? is the unit 

c 

stress at a unit's distance from the axis and _^?/ is the 

c 
unit stress at any distance y. 

Considering the stress uniformly distributed over any 

small area dA at the distance ?/, — ^dA is the force acting 

c 

on that elementary fiber, and — ? ^'^dA is the moment of 

c 

this force about the axis of the bar. But the sum of the 
moments of the forces acting in the section is equal to the 

twisting moment Pp, hence -^ / y^dA = Pp. I yMA is 



86 MECHANICS OF MATERIALS 

the expression for the polar moment of inertia about an 
axis through the center of gravity of the section, and 
writing J for „ ^ ^ 

I y^dA gives Pp = — , {e) 

which shows the relation between the maximum unit stress 
in shear and the twisting forces. 

Art. 53. Modulus of Section. 

Comparing formula (e) with M— — , it will be noticed 

c 

that they are of the same general form, ilf and Pp repre- 
sent the effect of the external forces, S in each case is the 
maximum unit stress in the section, and by analogy with 

-, - may be called a modulus of the section. Formula 
c c 

(e) is only true for bars whose sections are circular, and 
where the material and loading of the bar satisfies the 

conditions stated for M = — 

c 

Art. 54. Square Sections. 

For rectangular sections the formula is only approxi- 
mately true, as a radial line drawn from the corner does 
not remain straight during the twisting of the bar, as was 
the case with the circular section. 

The investigations of St. Venant cover the rectangular 
sectioli, and his results give for a rectangular shaft sub- 
jected to torsion, g 

■^P^ ^'^ (nearly), 

in which d is the side of the square. From the form of 

J d^ 

the expression the effective value of - must be --— instead 
^ G 4.8 



TORSION 87 

of the calculated value — - — , hence the value of Pp as 

6 

found from the above formula is less than that obtained 
from equation (e), when equal values of S^ and d are 
used. 

Square sections are apt to be weaker than St. Venant's 
formula would indicate, since the maximum stress is 
carried on the edge of the square and any slight defect 
reduces the effective diameter of the bar. 

For this reason square sections are rarely used to resist 
torsion alone. 

Art. 55. Illustrations. 

The resistance at the wall may be assumed as another 
couple, whose moment is equal and opposite to that of 
the twisting moment, without altering the conditions as 
assumed when formula (e) was developed. In the case 
of a line shaft, where the belt from the engine produces 
a twisting moment at one end of the shaft, the resistance 
of the belt on the pulley at the other end is equivalent 
to an equal moment. If there are several pulleys on the 
same shaft, each producing a moment by its resistance 
to turning, it is evident that the resisting moment of the 
shaft will not be constant throughout the length, but will 
vary with the resistance that it has to overcome. To 
illustrate, assume a shaft with three pulleys, one at each 
end and one at the middle of the shaft. The driving 
moment is applied at the left end and the resisting 
moment of the other two pulleys, P^Pi and P^p^-, must be 
equal to the driving moment Pp. The moment to be 
resisted by the shaft between the driving pulley and the 
one at the middle is Pp = P^fx + ^ilPv After passing 



88 MECHANICS OF MATERIALS 

the middle pulley the resistance to be transmitted by the 
shaft is only the twisting force of the third pulley, and 
consequently the resisting moment between the second 
and third pulleys is equal to P^ip^,' 

Art. 56. Twist of Shafts. 

In Fig. 52 Z», dh was taken as a measure of the defor- 

mation of the surface fiber, and — the unit deformation 

of that fiber. From the figure dh = 6c and by the defi- 
nition of the modulus of elasticity, if S^ is the unit stress 
in the surface fibers, the shearing modulus of elasticity 

i^^^st be Q ^n T> 1 

^_S,_SJ. Ppl 

I 

The latter expression is obtained by substituting for S^ 
its value from /S'„ = —^- 

Equation (/) gives the relation between the modulus 
of elasticity for shear, the twisting moment or the unit 
shearing stress, and the angle of twist. 

When the data given in any problem is sufficient to 
determine two of the three quantities, the twisting moment, 
the unit shearing stress, or the section modulus, the 

cr J 

formula Pp=— ^— will completely determine the other 
c 

one. Or when the given data will determine three of 
the following quantities, the twisting moment or the unit 
shearing stress, the dimensions of the bar, the angle of 
twist, and the shearing modulus of elasticity, formula (/) 
can be used to solve any problem involving the twisting 
moment or the unit stress and the angle of twist. 



TORSION 89 

Since all of the tabulated values of the constants of 
materials are given in pounds and inches, all dimensions 
of weight and linear or square measure must be reduced 
to pounds and inches before making the substitutions in 
the formulas. As the value of 6 used in the development 
of the formula was in circular measure, 6 must be ex- 
pressed in radians. 

Art. 57. Relative Strengths and Stiffness of Shafts. 

The strength of a shaft may be defined as the twisting, 
moment it will carry with a given unit stress. Formula 
(e) shows that the strengths of shafts vary directly as 

the value of -^— for each shaft, and when the shafts are 

^ . J 

of the same material, as — 

c 

Defining the stiffness of a shaft as the angle of twist 

for a given value of P^, since = -^ or _^, it is evident 

that the stiffness of two shafts of the same material varies 

directly as — or -, depending on whether the twisting 

force is given directly or in terms of the stress. 

Art. 58. Horse Power of Shafts. 

A horse power being defined as 33,000 ft. -lb. per 
minute, if H is the horse power to be delivered by a shaft 
making N revolutions per minute, the value of Pp in 
terms of the horse power may be found from the equality 
of the work done by the twisting force per minute and 
the work represented by J?^ horse powers. Assuming that 
J^ is a force acting at a radius jt?, the work done by that 
force in one revolution must be 2'TrPp in. -lb. and as 
a horse power is 33,000 ft. -lb. per minute, or 396,000 



90 



MECHANICS OF MATERIALS 



in. -lb. per minute, 2 TrPpiV^ = 396,000 H, wliicli re- 

PpN 

duces to ■g"= /^ (approximately). Substituting for 

Pp its value in terms of S^ from formula (e), gives 
jy= ^t) ^rvr. — These two expressions may be used to de- 
termine the horse power that a given shaft will transmit 
when the number of revolutions per minute and the twist- 
ing moment or the maximum unit stress are known. 
The values of either Pp or Sg may be found from (/) 
and the angle of twist for a given horse power determined. 

Art. 59. Shaft Couplings. 

When two lengths of shafting are to be joined together, 
the connection is often made as in Fig. 59. The moment 





Fig. 59. 

of the shearing stresses in the bolts must be equal to the 
twisting moment of the shaft, and the relation between 
the twisting moment of the shaft and the resisting 

moment of the bolts may be stated as Pp = SJ — —^ 

where J' is the polar moment of inertia of the section of 
a bolt about the axis of the shaft, n the number of bolts, 
SJ the maximum unit stress in the bolts, and <?' the dis- 
tance of the most distant fiber of the bolt from the 
axis of the shaft. 



TORSION 91 

If the bolts and shaft are of the same material, then 

- = — - , where J and c refer to the shaft and J' and c' to 
c c 

the bolts. The polar moment of inertia of the bolt about 
the axis of the shaft is equal to J"' = J"^ -f Al^^ where Jg 
is the polar moment of the bolt section about an axis 
through its center of gravity, and A is the distance of the 
axis of the bolt to the axis of the shaft. Expressing 

_ and - — - in terms of the diameters of the shaft and 
c e' 

bolts, gives a method for finding the proper diameter of 
the bolts to be used. The equation thus formed gives a 
very awkward expression for the value of the diameter 
of the bolts, and it is common practice to assume that the 
shearing stress is uniformly distributed over the area of 
the bolts. This is equivalent to assuming that the result- 
ant of the shearing stresses acts at the center of the bolts. 
Let SJ^ be the uniform unit shearing stress in the bolts, 
h the radius of the bolt circle, or the distance between the 
center of the shaft and the center of the bolts, and d the 
diameter of the bolts; then the resisting moment of 

7rc?2 
the bolts must be n—^SJ' A, and equating this to the 

twisting moment of the shaft gives 

which is in convenient form for use in the determination 
of the bolt diameters. 

As the resisting moment of the bolts must be equal to 
the resisting moment of the shaft, if Sg is the maximum 
unit stress in the shaft and I) the diameter, then 

G 4 * 



92 MECHANICS OF MATERIALS 

gives the relation between the unit shearing stresses in the 
shafts and bolts. Either the number of the bolts, their 
diameter, or the radius of the bolt circle can be taken as 
unknown. 

In applying the approximate solution to the solution of 

a problem, the total bolt area, i^~r~^ needed, may be found 

and the number of bolts chosen that will give the required 
area. The assumption made in the approximate solution 

is also equivalent to assuming that J' = — - — and c' = h. 

If h = ad, the unit stress derived by the use of the approxi- 

4a— 1 
mate formula is — - — -, 100% less than the unit stress 

8a^+l 

obtained by the use of the theoretical expression. For 

usual values of a, this reduces to -— — - nearly. The error, 

2a 

which decreases as the ratio oi h to d increases and there- 
fore not of much importance in the majority of cases, 
should always be considered when decision is made regard- 
ing the bolt diameters, as the total bolt area determined by 
the use of the approximate formula is less than the area 
required by the more exact theory. 

Art. 60. Modulus of Rupture in Torsion. 

When the twisting moment Pp is great enough to rup- 
ture the bar, the value of S^ as found by equating this 

value of J*p to — has been called the modulus of rupture 

in torsion. The value of such a constant of material is 
doubtful, as the formula is not true when the unit stress 
is greater than the elastic limit. It would seem to be 
better practice to use the maximum unit stress in shear, 
as determined by the use of P = AS as the maximum unit 
stress in torsion, and base any factor of safety on that 
constant. 



TORSION 



93 



Art. 61. Helical Springs. 

When a wire is wound around a cylinder so that the 
axis of the wire forms a hehx, the resulting form of the 
wire is called a hehcal spring. Given a helical spring 
having a diameter of wire, 
d, and the mean diameter 



of the coil, D. Assume 
the spring to be com- j" 
pressed or extended by a ^ 



AP 





force P, whose line of ac- 
tion passes through the 
centers of all the coils and 
consider one coil. If the 
spring is closely wound Fm. 61. 

and D is large compared 

with d, the plane of the coil will be nearly perpendicular 
to the line of action of the force; hence the force P acting 

D 
with a lever arm — tends to twist the wire. 

PD 

The twisting moment is — --— and the section modulus 

of the wire tt— -; hence, taking* iS, as the unit stress = 

16 ^ ' 2 

u /-v ird 

IT — S,, or P = — — - S, g-ives the relation between the ten- 
16 8 i> ^ 

sile or compressive force P and the dimensions of the 
spring. 

The length of one coil of the spring is approximately 
ttD, and if the wire is twisted through a small angle 0, by 



a moment that increases uniformly from to —^^ the 

PD 

work done on the wire of one coil is 6 — ■ — , where 9 is in 

1 

circular measure. If we let A be the total deflection of 
the spring, that is the amount of shortening or lengthening, 



94 MECHANICS OF MATERIALS 

and B the deflection for one coil, as the force acting varied 

uniformly from to P, the work done on one coil is 

Since these two expressions represent the same quantity 
of work, they must be equal, or = — — or = — . 

As 6 is the angle of twist, formula (/) gives as its value 

FJ Fd^ ' 
Equating the two values of ^, 

lb - — — = — or = , 

Fd^ D Fd^ ' 

which is the deflection of one coil in terms of the load P. 

Taking • e=^=~ 

Fc i>' 

we have h = — ^— — • as the relation between the unit stress 
Fd 

and the deflection for one coil. 

Since the strength of any bar under the action of twist- 
ing forces is independent of the length of the bar, the 
formula for the strength of a spring is also independent of 
the number of coils. If n is the number of coils, the total 
deflection of any spring must be n times the deflection for 

one coil, or ^^^3^ ^^^^2^ 

A = or -^ ■• 

Fd"^ Fd 

EXAMINATION 

1. What is a torsional force ? How does it differ from 
a force that produces a shearing stress that is uniform over 
the section of a bar ? 

2. Can a torsional force produce any tensile or com- 
pressive stresses ? 



TORSION 95 

3. " It is evident that the angle had is independent of 
the length of the bar." Prove it. 

4. Show that the angle of twist depends on the length 
of the bar. 

5. If a straight radial line drawn on the end of a bar 
remains straight when the bar is being twisted, show that 
this fact proves that the unit stress is proportional to the 
distance from the axis. 

6. Show that Ss=-^r-- 

7. State under what conditions of load and material 

the formula Pp = — ^— is true. 

c 

8. Define the terms "strength of a shaft"; "stiffness 
of a shaft." 

9. Does the strength of a shaft depend on its length ? 
Is the same true of the stiffness ? 

10. Two shafts of the same diameter and length are of 
different materials. What is their relative strength ? 
What is their relative stiffness ? 

11. Show how to find the expression for the modulus of 
elasticity in shear. 

12. Why was it necessary to reduce the value of a horse 
power, 33,000 ft. -lb., to inch-pounds in order to obtain the 
expression SJJST ^ 



63,000 c' 

PROBLEMS 

1. One end of a circular bar 2 in. in diameter and 10 ft. 

long is fixed in a wall, and at the other end there is a 

couple whose moment is 300 ft.-lb. Required the unit 

stress in the bar ? 

Solution. The value of J = ^ — and c = — ; hence - = — — . 
Pp = 300 ft.-lb. = 3600 in.-lb. ^^ . ^ c 16 



96 MECHANICS OF MATERIALS 

S J 

Substituting these values in Pp = - — , and solving for Ss, gives 

c 

o 3600 X 16 nnnrv ii, / 

Ss = = 2290 Ib./sq. m. 

TT X 8 

2. A circular bar 7 in. in diameter, 10 ft. long, is acted 
on by a force of 10 tons perpendicular to, and at a distance 
of 3.14 ft. from tlie axis. Required the unit stress induced. 

3. Find the diameter of a circular steel bar to carry a 
twisting moment of 20 ft. -tons. 

4. Find load that can be applied at the end of an arm 6 
ft. long so that the maximum unit stress induced in a cir- 
cular bar 2 in. in diameter will not exceed 12,000 Ib./sq. in. 

5. A circular steel bar 2 in. in diameter is twisted by a 
force at the end of an arm 6 ft. long. Required the force if 
the unit stress due to torsion is equal to the elastic limit. 

6. If the bar in problem l was soft steel, find the angle 
of twist. 

Solution. Taking the values as found for 1 and substituting in 

= ^, Z rr 10 X 12 in. and F = 12,000,000, 
Fc 

^ ^ 2290 X 10 X 12 ^ 22,900 ^ ^229 radian, 

12,000,000 X 1 1,000,000 

or approximately 1° 18'. 

7. A soft steel bar is 6 in. in diameter and 20 ft. long. 
What force acting tangent to the surface will twist the 
bar through an angle of 1° ? 

8. A steel shaft 2 in. in diameter and 50 ft. long is 
transmitting a torsional moment that causes a unit stress 
equal to the elastic limit. Required the angle one end 
is twisted through relative to the other. 

9. Find the diameter of a steel shaft 10 ft. long to 
carry a twisting moment of 81,700 ft. -lb., if the unit 
stress is not to be greater than 10,000 Ib./sq. in., and 
the angle of twist less than 1.15°. 



TORSION 97 

10. Find the diameter of a steel shaft making 100 
revs. /inin. and transmitting 200 H.P., the unit stress 
being 6300 lb. /sq. in. 

S,JN 6300 X TT X c?3 X 1 00 nnn 
Solution. ^ = 0^^= 63,000x16 "^^^^ - 

^3 ^ 200 X 63,000 X 16 ^ ^ g^ .^^ 
6300 X TT X 100 

The shaft chosen would be either 4^ or 5 in. in diameter. 

11. The Allis Chalmers Co. base their tables for the 
strength of mild steel shafting on the formula J£= cd^ JV, 
where d is the diameter of the shaft in inches, and c a 
number which has the following values: 

For heavy or main shafts c = .008 

For shafts carrying gears c= .010 

For light shafts carrying pulleys c = .013 

Find the unit working stress allowable in each case. 

12. A hollow steel shaft is 12 in. external and 10 in. 
internal diameter. Find the H.P. that may be trans- 
mitted at 100 R.P.M. when the unit stress due to torsion 
does not exceed 6000 Ib./sq. in. 

13. A flange coupling is to be used to connect two lengths 
of 6 in. steel shafting. The maximum allowable shearing 
unit stress in the shaft and bolts is 6000 lb. /sq. in. Assume 
the diameter of the bolt circle to be 8 in. and find: 

(a) By the use of the approximate formula the total 
bolt area required. 

(6) Assuming the number of bolts as 6, determine the 
true unit stress in the bolts. 

14. A hollow shaft has the outside diameter twice the 
inside diameter. Compare its strength with that of a 
solid shaft of the same material and section area. 

15. If the elastic limit of the material in one shaft is 
60,000 lb. / sq. in. and costs 10 ^ per pound, what can 



98 MECHANICS OF MATERIALS 

you afford to pay for a shaft to do the same work, if the 
elastic limit of the material is*30,000 lb. /sq. in. ? 

16. A helical spring is made of wire whose diameter is 
1 in., the mean diameter of the coil 4 in., and has 30 coils. 
If the value of F is 12,000,000 and the working unit 
stress 60,000 lb. /sq. in., required the load it will carry 
and the deflection under that load. 

17. Find the size of the wire and the mean diameter of 
the coils for a steel helical spring with 30 coils to carry a 
maximum load of 6000 lb., have a deflection of 10 in. under 
that load and a limiting unit stress of 60,000 Ib./sq. in. 

18. D. K. Clark gives as the deflection for one coil of a 
helical spring h— ^^ , where d is the diameter of the 
wire in sixteenths of an inch. 



CHAPTER V 



THE ELASTIC CURVE 

Article 62. Definition. 

When the beam is bent, the neutral surface assumes 
a curved form; and the projection of this surface on a 
vertical plane parallel to the axis of the beam is called 
the Elastic Curve. If the equation of this curve for any 
beam is expressed as y^f(x), where y is the deflection 
of the beam at any point distant x from the left end of 
the beam, the deflection of the beam at any point can be 
easily found. 

Art. 63. Equation of the Elastic Curve. 

To derive this 
equation of the 
elastic curve let 
Fig. 63 represent 
a portion of a bent 
beam. 

Let ji measured 
along the axis of 
the beam be tak- 
en as representing 
dl., where I is the 
length of the beam 
and the curve dl a 
differential part of 




Fig. 63. 



99 



100 MECHANICS OF MATERIALS 

the elastic curve, then if o is the center of curvature, oi 
and oj are radii. Before any bending took place these radii 
were parallel to each other, and eh may be assumed to have 
been in the position U parallel to af. Let the greatest 
distance of the neutral axis to the surface fiber be c and 
assume that jh = c. The deformation of the outer fiber 

is kb and the unit deformation is — , since Jcb is the change 

dl ^ 

in the length dl; J^ is the modultis of elasticity, and S 

the unit stress in that fiber, hence kb = ^-—-. From the 
similar triangles jkb and oij 

— = ^. ri) 

ji of 

But oj = r = the radius of curvature, ji = c?Z, jb = c, and 

kb — -^7- ; hence, substituting these values in (1), we have 

Sdl c ^ ^ \^ , CI -^^ 
Edl r E r I 

. M E El ,oN 

hence -_ = — , or r = — -, (2) 

I r M ^ ^ 

is the equation of the elastic curve of a beam in terms 
of the modulus of elasticity, the bending moment at any 
point of the beam, the moment of inertia, and the radius 
of curvature of the elastic curve at that point. 

The value of r expressed in rectangular coordinates is 



1+r^ 

'•^ § ^^^ 

dx^ 

Since the degree of curvature of any beam in an en- 
gineering structure is very small, the value of the tangent 



THE ELASTIC CURVE 101 

of the angle which the tangent at any point on the curve 
makes with the axis of ^ is a very small quantity and 

-^ j may be neglected in comparison with unity. Equa- 
tion (3) then reduces to 

If dx is assumed to be equal to dl^ an assumption which 
is approximately true when the degree of curvature is 
small, and the value of r as just found is inserted in 
equation (2), we have 

dx^ 

which is the differential equation of the elastic curve of 
any beam. M is the bending moment for any part of 
the beam for which the equation represents the curve, 
therefore must be expressed in terms of x any distance 
from the left end of the beam, and y is the ordinate to 
the curve or the deflection of the beam. 

Since the condition that jE^ = — was introduced in the 

€ 

derivation, the unit stress must be within the elastic limit, 

Mc 

and as the formula aS' = -— was also used, all the condi- 
tions of the materials that are necessary to the correct 
use of that formula obtain with the one just developed. 
The assumptions that dl = dx^ and that the maximum 

value of -^ was so small that ( — ) might be neglected as 
dx \dxj 

compared to unity, introduce errors that, while they are 

small, increase as the degree of curvature increases. The 



102 MECHANICS OF MATERIALS 

use of the formula under ordinary engineering conditions 
gives results that agree well with those obtained in exper- 
imental work, the discussion of the limitations to its use 
being given to show that the formula is not applicable to 
all cases. 

Art. 64. Deflection of Beams. 

Let M = f(x) be the expression for the bending moment 
for a portion of any beam loaded in any way; then 

mfi=fix) (1) 

is the differential equation of the elastic curve for that 
part of the beam for which f(x) represents the bending 
moment. 

Integrating (1), 

EI^=fix-)+0, (2) 

where (7 is a constant of integration. The value of C 
may generally be found by noting that -^ is the tangent 

of the angle which the tangent at x makes with the axis 
of X, and from the conditions of the problem, finding a 

value of X for which -^ is either zero or known. 

dx 

Integrating again, 

EIy=f"(ix')^-Ox+ C,. (3) 

The value of (7^, the constant of integration, can in 
most cases be determined by finding a value for x for 
which 1/ is either zero or known. 

When O and C^ are determined, equation (3) may be 
used to find the deflection at any point of the beam for 
which the bending moment is equal to /(a;). 



THE ELASTIC CURVE 103 

As EI— is the differential coefficient of Ely, the value of 
dx 

dy 
X that makes EI— = will determine the maximum value 

dx 

of y, provided that this value of x falls within the limiting 
values of x, for which f{x) =M, in which case 2/ = maximum 
deflection. 

In the determination of the constants of integration, 
the student is reminded that when the moment diagram 
is symmetrical about a vertical line at the middle of the 

beam, — is zero for a value of a; =— , and for the special 
dx 2 

dv 
case of cantilever beams ~ is zero at the wall, as the re- 

dx 

strainment keeps that part of the beam fixed and horizontal. 

Since there is no deflection at the supports, 1/ will always 
be zero at the points where the beam is supported. When 
there are concentrated loads on the beam, the expression 
for itfwill take a different form for each part of the beam 
between the loads or between the loads and reactions. 

If there are n concentrated loads on the beam, there 

will he n -\- 1 forms that the expression for the bending 

moment may take, making n -\- 1 equations of the elastic 

curve, tlie double integration of each bringing into the 

problem 2 (^ + 1) constants of integration. 

For simple beams the value of x that will make -^ = 

dx 

is not known unless the loads are symmetrical with the 
middle of the beam, but ^ is always equal to zero at the 
supports where x is either or I. 

Any two of the 7^ + 1 equations representing consecu- 
tive portions of the beam will have the expressions for — ^ 

dx 

and i/ equal for a value of x at the load where the equa- 
tions meet. As there are but 2 n -\- 2 constants of inte- 



104 MECHANICS OF MATERIALS 

gration, and two known conditions are always to be had 
from the fact that ?/ = when x= or x = I, the 2 n 
equations resulting from the equating of the values of 

— ^ and y under each load will suffice to determine the 
dx ^ 

other 2 n constants of integration. 

In using equation (3) for the determination of the 
maximum deflection of a beam, the student is again 
reminded that the equation only holds true for tliat 
portion of the beam represented by /(a;), and where 
there are several concentrated loads, each portion may 
have to be investigated, in order to find the greatest 
value of ^ for the beam. In general, an inspection of 
the distribution of the loads on the beam will show the 
portion of the beam where the greatest deflection is likely 
to occur. 

If the bending moments for simple and cantilever beams 
loaded uniformly with TFJ simple beams loaded at the 
middle with TF, and cantilever beams loaded with TTat the 
end, are expressed in terms of x^ these moments may be 

substituted in UI —^ = iHf, forming four differential equa- 
dx^ 

tions. Integrating each equation twice gives the value of 

y, the deflection of the beam for any value of x. The 

maximum value of the deflection y may be expressed as 

1 WP 
A= — — --, where /3 is a constant depending on the kind 

of a beam and the nature of the loads. Transposing, 

W= /6— -— . The load a similar beam will carry was 

ST 

given as TF= a — -, and if these two expressions for W 

^^ a SP 

are equated we find that A = — — -, giving the relation 



THE ELASTIC CURVE 105 

between the maximum deflection, the dimensions of the 
beam, the unit stress, and the modulus of elasticity. 

Attention is called to the way A varies in the two 
expressions for the maximum deflection. 

When the load W is considered, A varies directly as P 
and inversely as Z, and when the unit stress is given, A 
varies directly as P and inversely as e. For rectangular 
sections, this latter variation makes the maximum deflec- 
tion independent of the breadth of the beam. 

Aet. 65. Restrained or Fixed Beams. 

A beam is said to be restrained or fixed when one or 
both ends are so firmly imbedded in the wall that the tan- 
gent to the elastic curve at the fixed ends always remains 
horizontal during the flexure of the beam. A cantilever 
beam under this definition is a fixed beam when it projects 
from the wall ; but since there was no reaction at the free 
end, the bending moments and shears can be determined 
without reference to the fixed end. When the free end of 
a cantilever beam has a support placed under it, the con- 
ditions are. changed. The magnitude of the reactions 
can not be determined from the conditions for mechanical 
equilibrium that were used for simple beams. The value 
of Jf for any section of the beam will contain a term that 
includes the unknown reaction, and this value of M sub- 
stituted in the general equation of the elastic curve will 
give the differential equation of the curve for this beam. 
When this equation is integrated twice, the value of the 
unknown reaction may be found by applying the known 
conditions that the resulting equations must satisfy. 

Let Fig. 65 represent a beam fixed at one end and 
supported at the other, loaded in any way. If B^ is the 



106 



MECHANICS OF MATERIALS 



reaction at the left end of the beam, the bending moment 
for a section distant x from that end is 



M= R^x - "ff- 



rthe sum of the moments of the loads' 
to the left of the section with refer- 
ence to a point in that section 



and putting this value of M\xv the general equation of the 
elastic curve gives 






( sum of the moments ) 
\ of the loads, etc. \ ' 



which is the general equation of the elastic curve for 
a beam fixed at one end and supported at the other. 

If there are no concen- 
^^^^ trated loads, the last 

zz^U/ l®^''^^ o^ ^^ right hand 

member will be zero, 

... and if there is no uni- 

will be 




form load, 



IVX" 



2 




Fig. 65. 



z^^:^^ zero. In a previous 
^^ article it was shown 

., that in a sfeneral case 

Wy'^////^. there were 2 (w + 1) 
conditions for the deter- 
mination of an equal 
number of integration constants, and in this case there is 

the additional condition that -^ = when x = 1, which 

ax 

may be used to determine i^^, since the tangent to the 
curve is horizontal at the wall. Stated briefly, the con- 
ditions that ?/ = at the fixed and supported ends, -^ == 
at the fixed end, and the ^n equations resulting from the 



THE ELASTIC CURVE 



107 



equating of the values of -^ and y under each load, will 

be sufficient to completely determine the values of the 
2 (n + 1) constants of integration, and that of the 
unknown reaction at the left end of the beam. 

Art. 66. Beams fixed at Both Ends. 

Let Fig. ^^ represent a beam fixed at both ends and 
loaded in any way. The forces which act between the 
beam and the walls supporting the beam, keeping the tan- 
gent to the elastic curve at the wall horizontal, are un- 
known. The un- 
known systems of '^'^^ 
forces acting in 
each wall may each A 
be replaced by a 
single vertical 
force acting up- 
ward at the face of 
each wall, and a 
couple whose mo- 
ment is sufficient 
to keep the tangent 
at the wall horizon- 
tal. The beam, 
under the action of 
these forces and the 
loads, may then be considered as a body in equilibrium. 
From the mechanics of equilibrium of parallel forces, it is 
evident that tlie sum of the two vertical forces acting up- 
ward must be equal to the sum of the loads on the beam, 
and if the moments of the couples are determined, the 
values of the two vertical forces may be found. 




Fig. 66. 



108 MECHANICS OF MATERIALS 

Let R^ (Fig. 66) be the force at the left end and R^ the 
force at the right end, and the moments of the couples at 
the left and right ends of the beam be M^ and M^^ respec- 
tively. Since the moment of a couple about any point is 
constant, the moment of the couple on the left end about 
a point in a section distant x from the left end is M^^ and 
the value of the bending moment for that section is 

■n/r n/r , -r> WOiP' (the suiii of the moments) 

J- ■■■ 2 (of the loads, etc. ) 

This value of iJf substituted in EI—^ — M'\^ the differ- 

ential equation of the elastic curve for a beam fixed at 
both ends. If there are no concentrated loads, the last 
term of the expression for M will be zero, and when 
there is no uniform load, the term containing w will 
disappear. If there are n concentrated loads, there will 
be ^ + 1 values for M^ and each expression will contain 
the unknown moment My The double integration of the 
71+1 equations will bring into the problem 2(n + 1) con- 
stants of integration, making 2 ti + 6 unknown constants 
to be determined, as My, M^, R^ and R^ are all unknown. 
Noting that R^ + R^ equals the sum of the loads, and 
that the moments of all the forces must be zero will give 

two, the values of y and — ^ being zero at each wall will 

dx ^ 

give four more, the derived values for y and -^ in the 

CbX 

adjacent expressions for M are equal for values of x 
under any concentrated load will supply 2 n equations, 
the required 2^ + 6 equations may be written. If the 
loads are symmetrical with the middle of the beam, the 

solution will be much simplified, as — ^ is zero at the 

dx 



THE ELASTIC CURVE 109 

middle of the beam, and R^ = R^^ and M^ = M^. The 
methods given are general and will completely determine 
the bending moments or deflections for any restrained 
beams. 

Having fully determined the expression for the values 
of M and V for any restrained beam, the shear and 
moment diagrams may be drawn, and the maximum mo- 
ments as well as the inflection points determined, as for 
simple beams. The value of x which makes the vertical 
shear equal zero gives the dangerous section for the beam, 

and the formula aS'= — with the value of ili'a maximum 

can be used for all investigations of the strength and 
safety, as well as the design of restrained beams. 

Art. 67. Continuous Beams. 

A continuous beam was defined as one having more 
than two supports. 

The supports are as- 1 00X^)^0000^^000 
sumed to be on the I Nil 



<-x-^ Uz 






same level and rigid, I ^_____.^ 

and the section of the VJ" 

beam uniform. ' Fig. 67. 

Let Fig. 67 represent any two intermediate spans of a 
continuous beam or girder whose lengths are l^ and l^ and 
loads per linear unit w-^ and w^ respectively. 

There is an unknown bending moment at each support. 
Let Ny, iV^, and iVg represent these moments. 

The moment N-^ is produced by the resultant of all the 
forces acting to the left of the first support. This result- 
ant and the reaction at the left-hand support may be re- 
placed by a couple whose moment is iV^ and a vertical 
force V^ acting at the left-hand support. Evidently F^ 
is the sum of the resultant force and the reaction at the 
left support. 



110 MECHANICS OF MATERIAL 

Similarly F^, and a couple whose moment is iV^g' ^^^^^ ^s 
and a couple whose moment is iVg, can replace the forces 
acting at and to the left of the second and third supports. 

Taking the origin at the left-hand support and x as any 
distance in the left-hand span from that support, the equa- 
tion of the elastic curve for that span becomes 

^jg=Jf = iVi+Fl^-^'. (1) 

Integrating twice and determining the constants by the 
known conditions that ^ = when x equals either or ?p 

. UI^^=W,x + ^-'^+0, (2) 

and m^=^^^-'^+0,x + C,. (3) 

(72=0 and 0^=-^ 2 6~ ' 

Substituting the value of 0-^ in (2), 

■^^^"■^i'^ + ~2 6~+^ 2 6~ ^^^ 

Taking the origin at the second support and x as any 
distance in the second span, the differential equation of 
the elastic curve for that span becomes 

^jj=i(f=iv, + r,^-^. (6) 

Integrating twice and determining constants from the 
known conditions that ?/ = when x equals either or l^^ 
v/e can write 



THE ELASTIC CURVE 111 

The values of -j- in (4) and (6) are equal when x=^l^ 
in (4) and in (6). (4) reduces to 

— -o-H Q 5—' \'J 



ax 



x=L 



and (6) to 



ax 



"^2^2 -^2^2 ^2^2 



:r=o 24 2 

From (1) M= N^ when x = ly. 



(8) 



Similarly^ from (5) M = N"^ when x — l^^ 

w 1 '^ TV — TV ?/; 7 

if=iV3 = iV2+F2?2-^. .•.r2 = ^^iy^2+^^ (10) 

2 

Equating (7) and (8) and substituting values of V^ 
and V^ from (9) and (10) and reducing, we have 

N,l, + 2 N, (I, + ?,) + N,l, = - ^^^'° + "'^^''' , (11) 

which gives the relation between the unknown bending 
moments iV^, iVg, and iVg, and the uniform loads w-^ and w^. 
One equation similar to (11) may be written for any three 
consecutive supports, and if n is the number of the sup- 
ports, n — 2 such equations may be written. 

When the ends of the beam are merely supported, the 
moments are iV^, iV2 • • • ^n-i ^^^ the values of iV^ and iV„ 
are each equal to zero, so that the n — 2 equations that 
may be written will completely determine the values of 
the 7^ — 2 unknown moments. 
' When both ends of the beam are to be fixed, the two 

additional conditions that -^ = at either end support. 



112 MECHANICS OF MATERIALS 

will furnish two more equations that will be sufficient to 
determine the unknown bending moments at the end 
supports. 

When the values of the bending moments at the sup- 
ports are determined, the maximum bending moment in 
any span may be found by writing for the desired span an 
equation similar to tliat given for M in (1). The forces 
Fj, 7^, and F^ are equal to the vertical shears just to the 
right of the first, second, and third supports respectivelyv 
Let R with subscripts 1, 2, and 3 represent the reactions 
at the corresponding supports. Then for the first support 

From the definition of vertical shear 

Fi - M>i?i + i?2 = Fj or iZ^ = Fj - ( Fj - w^l{). (12) 

As an equation similar to (12) can be written for any 

span, the reaction at any support can be determined. 

When there are concentrated loads on each span, the 

discussion is complicated by the fact that there are two or 

more forms for the equation of the elastic curve for any 

span instead of only one, bringing into the problem two 

constants of integration for each concentrated load. 

clii 
Notinor the additional conditions; that the values of -^ 

and y^ where two equations meet at the point of application 
of any concentrated load are equal for the value of x at 
that point, it is simply a question of algebra to find the 
relation between the three moments. 

The expression as given in equation (11) is known as 
the Theorem of Three Moments and was published in 1857. 

Knowing the reactions and the loads, the shear and 
moment diagrams may be drawn as for simple beams. 



THE ELASTIC CURVE US 

The maximum bending and shearing stresses may be 

found by using the maximum values of M and V in the 

Mc V 

fundamental formulas S= — — and S=—. 

I A 

The maximum bending moments occur where the shear 
passes through zero; hence, by writing the expression for 
the vertical shear and equating it to zero, the position of 
the maximum moment may be found and its magnitude 
determined from the equation for the bending moment. 
Equating the expression for the bending moment to zero 
will give the point of inflection. The caution given before 
may well be repeated here : 

" When substituting the value of x which renders the verti- 
cal shear zero to determine the value of the maximum bending 
moment^ substitute in the particular expression for the value 
of M that applies to that portion of the beam^ and when 
equating the expression for M to zero in order to find the in- 
flection points^ the expression which applies to that portion 
of the beam must be used. ^^ 

As w^s the case with overhanging beams, the inflection 
points and the points of zero shear may be approximately 
located by inspection of the moment and shear diagrams. 

The maximum stress in a continuous beam with a num- 
ber of equal spans may be less than the maximum stress 
for simple beams for the same spans ; hence, when using 
continuous beams care must be taken to insure that all 
the supports are on the same level and practically rigid, 
otherwise the maximum stress may be greater than for 
a number of simple beams. Take the case of two equal 
spans with a uniform load of w lb. /ft. If the beam is 
truly continuous, that is the supports on the same level 
and practically rigid, the maximum unit stress is, — 4 wl'^. 



114 MECHANICS OF MATERIALS 

This is numerically equal to the maximum moment in a 
simple beam of the same span. 

Suppose the middle support of the continuous beam to 
sink so that there is no reaction, then the maximum 
moment is ^ w 4:P = ^ wl?^ instead of \ wl\ or the unit 
stress will be four times as large as for the simple beam. 

As the deflection of beams as used in engineering 
structures is a very small quantity, the stress may easily 
approximate this maximum value when the supports do 
not remain precisely at the same level. 

EXAMINATION 

What is meant by the expression, "the elastic curve of a 
beam " ? 

Derive the differential equation for the elastic curve of 
any beam. 

Name all the conditions that must be satisfied if the 
use of the equation will give results that are approximately 
correct. 

What is the " deflection " of a beam ? 

Show how the differential equation of the elastic curve 
may be used to determine the maximum deflection for any 
beam. 

Prove that if the loading of any simple beam is symmet- 
rical with the middle of the beam, -^ = at that point. 

dx 

Why is it that when there are n concentrated loads on 
a beam, the expression for the bending moment may take 
n-\-\ different forms ? 

Show that the maximum deflections of simple and canti- 
lever beams rectangular in section, and uniformly loaded, 

vary as — and ^3. 

What is a restrained beam ? 



THE ELASTIC CURVE 115 

Why is it that the laws of mechanical equilibrium can 
not be used to determine the reactions for restrained 
beams ? 

Explain how the differential equation of the elastic 
curve may be used as an aid in the determination of the 
reactions for restrained beams. 

Explain how to find the maximum bending moment 
for any restrained beam. 

Will the maximum bending moment and the maximum 
deflection always be found at the same point in the beam ? 

State under what conditions they will be found at the, 
same point. 

What is a continuous beam ? 

State the various steps in the method used in deriving 
the equation known as " the equation of three moments." 

Write "the three-moment equation" for a continuous 
beam with equal spans and a uniform load on each span. 

Show how to find the reactions for a continuous beam 
after the equation of three moments has been found. 

Show that in the case of a continuous beam uniformly 
loaded the maximum unit stress may accidentally be greater 
than the maximum unit stress for a number of simple 
beams, one for each span. 

When the size of a beam for a given span has been 
found on the assumption that the beam was to be fixed at 
both ends, it is necessary to take precautions to preserve 
the restrainment. Why ? 

Show that in any uniformly loaded beam, simple, 
restrained, or continuous, if Vis the vertical shear on the 
right of any left hand support, and iV is the bending 
moment at that support, the maximum bending moment 

in that span is J^Max. = ^ + -^ — , in which w is the uniform 
load per inch. 



116 MECHANICS OF MATERIALS 



PROBLEMS 

' 1. Show that if the depth of a rectangular beam of 
uniform strength is constant, the elastic curve is a circle. 

Solution. We have the relation _- = £, fro!n which r is constant 

E r 

when c is constant, but c = — ; hence, r is constant and the curve a 

circle. 

2. Show that if a simple beam carries two equal loads 
at equal distances from either end of the beam, the elastic 
curve between the loads is a circle. 

3. A simple beam 80 ft. long carries a uniform load 
of 160 lb. /ft. The beam is 4 in. wide and 6 in. deep. 
Modulus of elasticity 1,600,000. Required, the inclina- 
tion of the beam with the horizontal at the supports. 

Solution. Find the value of -H from the differential equation 

dx 

of the elastic curve ; put x = and solve. 

4. Find the maximum deflection for a simple beam, 
length Z, modulus of elasticity E^ moment of inertia Z, 
loaded with 

(a) a uniform load of w lb. /in. 

(5) a single load P at the middle. 

Solution for (a). The expression for the bending moment 

at any section of the beam is M = R^x and Ri = — ; hence 

EI — - — — • is the differential equation of the elastic curve for 

dx'' 2 2 ^ 

this beam. 
Integrating, 

£:/ ^ = ^' - ^' + C; but^ = when x = l, therefore C = - 4'; 
dx 4: Q ' dx 2 24' 

J u i-i. *.♦ -i- 1 -r-i dy wlx^ wx^ wl^ 
and substituting^ its value, hi -^ = . 

dx 4 6 24 

Integrating again, 

ETy='^- — -'^+ 0,; but as v = when x = 0, C. = 0, and 
■^ 12 24 24 ^' ^ ' 1 ' 

the equation of the elastic curve becomes 24 Ely = 2 wlx^ — wx^ —wl^x. 



THE ELASTIC CURVE 117 

Equating the expression for -^ to zero, we find that x = -, — .365 /, 

cix ^ 

and 1.365 l. The latter values have no meaning for this problem, and 

y is a maximum when x —-. Substituting this value for x in the 
expression will give the value of the maxinmm deflection. 

5. Find the maximum deflection of a cantilever beam, 
length I inches, modulus of elasticity ^, moment of 
inertia Z, loaded with 

(a) a uniform load of w lb. /in. 
(6) a single load at the end. 

Solve by taking the origin at the wall and measuring x toward the 
free end, and also at the free end of the beam. 

(c) a single load P, at a distance M from the wall. 
Suggestion for (c). For any section on the left of P, M = 0; 

hence that part of the beam is straight. The value of -^ for x = kl 

(Ix 

gives the slope of the straight portion. When the deflection under 
the load is known, the deflection of the end of the beam may easily 
be found. Take the origin at the wall. 

6. A steel cantilever beam rectangular in section is 
18 in. in length, and is to carry a load of 1000 lb. at 
the free .end, and deflect .36 in. under that load. The 
unit stress may be taken as 30,000 Ib./sq. in. Find 
the size of the beam. 

7. A beam fixed at one end and supported at the other 
is I in. long and carries, 

(^a) a uniform load of w lb. /in. 
(h) a single load P at the middle ; 
find the expressions for, 

(1) the deflection at any point of the beam. 

(2) the maximum deflection and the point where it 
occurs, 

(3) the maximum bending moment and where it occurs, 

(4) the reaction at the supported end of the beam, 
and make a sketch showing the form of the shear and 
moment diagrams. 



118 MECHANICS OF iMATERIALS 

Solution for (a). Taking the origin at the supported end let 
R^ be the unknown reaction at the supported end. The value of 

M at any point of the beam is Af= R^x , as there are no con- 
centrated loads on the beam, and the differential equation of the 
elastic curve becomes 

Ei'^.^n..--4. (1) 

Integrating, ij/ 1 = ^' _ !|! + c. (2) 

dy 
The tangent at the fixed end is zero, hence -^ = when x = 1, and 

substituting x=^l in (2) gives C = -^ ^. Substitute this value 

of C in (2) and integrate, 

^- R,x^ wx^ wPx Rd'^x ^ .„. 

This equation must be true for a; = when y = 0, therefore C^ = 0. 
It must also be true for y = and x = l; making y = and x = I, 
(3) becomes 

Substituting for Cj and R^ in (3), we have 

48 Ely = 3 ivlx^ - 2 wx^ - wPx (3') 

as the equation of the elastic curve for this beam. 

The value of M was R,x , and as it, = , 

~ 8 T"' ^ ^ 

also V = R^ — tvx = tvxy (5) 

8 

and from these two equations the shear and moment diagrams may 

be drawn. 

3 I 
V = for X — — ; therefore the maximum moment occurs at | Z, 

8 

and this value of x substituted in (4) gives 

M - ^ ^^' 



THE ELASTIC CURVE 119 

I'here will be a negative moment at the wall, and its value can be 
obtained by making x = I in (4), or 

*-' 8 

Making M = and solving (4) for x gives 

,^ 3 irlx wx^ r. 3 ? 

31 = = 0, or X = — , 

8 2' 4 

as the inflection point. 

Substituting for C and R^ in equation (2) gives 

■pT^IjL — ^ it'/a:^ ^ijx^ tvl^ 3 wl^ 

This expression is the differential coefficient of (3) ; therefore, 
equating the right hand member to zero and solving for x gives the 
value of X for which (3) is a maximum. This results in a cubic 
equation, 8 x^ — 9 Ix^ -{- P = 0. The roots of this equation are x = I, 
.42 Z, and — .298/. The latter value has no meaning for this beam, 
and X = I is a minimum; therefore the maximum deflection occurs 
at X = .42 l. 

Substituting x, .42 I in (3') gives 2/max. = as the maximum 

deflection. 

Suggestion for (&). There are two differential equations for 
the elastic curve and the two curves have a common tangent and 

common deflection for a value of a; = -. 

2 

8. A beam I in. long is fixed at both ends and carries, 
(a) a uniform load of w lb. /in., 
(6) a single load P at the middle ; 
find 

(1) the expression for the deflection at any point of 
the beam, 

(2) the expression for the maximum bending moment 
and its position, 

(3) the expression for the maximum deflection and 
where it occurs. 

(4) Make a sketcli showing the form of the shear and 
moment diagrams. 



120 MECHANICS OF MATERIALS 

Suggestion for (b). Since the loads are symmetrical with the 

middle of the beam M^ = M^ and i2j = i?2 = "o' ^"^ TT^ ~ ^> when 

X = I, X = —, and x = 0. 

2 

9. A rectangular beam, 20 ft. long fixed at both ends 
carries a uniform load of 100 lb. /ft. If the modulus of 
elasticity is 1,500,000, the safe unit stress is 500 Ib./sq. in. 
and d = 4:b, find the size of the beam. 

10. If the beam in 9 carried a load of 2000 lb. at the 
middle, and the other data the same, find the size of the 
beam. 

11. Find the maximum deflections for the beams in 
problems 9 and 10. 

12. Draw the shear and moment diagrams for the 
beams as given in problems 9 and 10. 

13. If the beams in problems 9 and 10 were to be beams 
of uniform strength w4th a constant depth, sketch the 
plan and elevation for each beam. 

14. Sketch the shear and moment diagrams for the 
beams in problem 13, using as ordinates the unit shearing 
stresses instead of the vertical shears, and the unit bending 
stress instead of the bending moments. 

15. Select a standard steel I beam to be used as a con- 
tinuous girder for four equal spans of 8 ft. each. The 
ends are simply supported and the beam is to carry a uni- 
form load of 7000 lb. / ft. Unit stress not greater than 
16,000 Ib./sq. in. 

Solution. As the ends are supported, N^ = N^ = and from the 
symmetry of the spans and loads, iVg = N^ and R^ = R^, R^ = R^. 
The three-moment equation for equal loads and spans reduces to 

Beginning with the second span, 

N,+ ^N, + N,= -'f, (2) 



THE ELASTIC CURVE 121 

and with the third span, 



Wi 



72 



N,^4.N,-\-N, = --^. (3) 

Since N2 = N^, (2) becomes 

2iY, + ^N, = -~, (4) 



AT - _f^ 



and as N^ = 0, eliminating N^ from (3) and (4), we find that 

IV ['■^ 
14 
Making iV^ = and substituting for iVg in (1), 

^ 28 

V 

For the first span, F = F, — ivx or V = when x = — 1. 

w 

The bending moment for any section of the first span [see equa- 
tion (1), art. 67] is 

M = N,-\- V,x - ^' and !/„,,,. = ^1 + |^. (5) 

Y 

Similarly, for the second span V = V^ — wx, or F= when x =—^, 

and as the value of the bending moment for any section in the second 
span is 

M=N,-\- V2X - ^, the iH,,ax. =N2 + ^. (6) 

2 2 w 

Equations (9) and (10) of Art. 67 give value of Vi and V^- Sub- 
stituting these values in (5) and (6), we have 

•^^max in the first span = -^^ tvP = approx. ^^^ ivP, 
^ 1568 ^^ 28 

-^^max in the second span = ivP = approx. — wl'^, 

and as the moments in the third and fourth spans are the same as in 
the second and first, the greatest moment occurs at the second sup- 
port and is —^ , which may be written ^ , where TF is the total 

28 28 

load on one span. 

Substituting this value of i^f in — = - , we find the value of - as 

S c c 

36 in.3. 

From the tables we find that a 12-in. I beam weighing 31.5 lb. /ft. 

has a value of - = 36, and that beam will be chosen. 
c 



122 MECHANICS OF MATERIALS 

16. I beams are to be chosen to cover two equal spans 
of 10 ft. each. The uniform load for each span is 5000 
lb. /ft. and the maximum allowable unit stress is 15,000 
lb. /sq. in. Choose a standard I beam on the assumption 
that the beam is continuous. What beam must be used 
if two simple beams are used instead ? 

17. Select a standard steel I beam to cover three spans 
of 8 ft. each and carry a uniform load of 7000 lb. /ft. The 
unit stress is not to exceed 16,000 Ib./sq. in. What size 
beam would it be necessary to use if each span was covered 
by a simple beam ? How much weight of steel will the 
use of continuous beams save ? 

18. In problem 17 find the reactions at the supports for 
each kind of a beam. 

19. Select a continuous steel I beam to cover two equal 
spans of 12 ft. each, and carry a load of 36,000 lb. at the 
middle of each span. Unit stress 12,000 Ib./sq. in. 

Suggestion. There are four forms of the differential equation 
of the elastic curve, each two having a common tangent and deflection 
at the load or reaction where they meet. The three reactions are all 
unknown but R^ = R^, and there is an unknown bending moment 
at R^., the middle reaction. 



CHAPTER VI 
LONG COLUMNS 

Art. 68. Stresses in Long Columns. 

A theoretical bar under the action of axial compression 
should always yield by crushing, since the resultant of 
the stresses and the axial loads are in the same line. In 
dealing with the actual materials and loads, the force 
due to the load is not always axial, no matter how much 
care is taken to make it so, and no material in common 
use is absolutely uniform. In nearly every case of bars 
under axial compression, there is more or less of a couple 
caused by the bar not being straight and uniform in struc- 
ture, or the loads not axially applied. 

The effect of this couple is to produce bending in the bar. 
As the amount of bending or deflection of a beam under 
the action of bending forces varies directly with the cube 
of th^ length, it is easy to see that the danger of bending 
a bar under the action of compressive forces which are not 
exactly axial, increases very rapidly with the length, and 
also that any bending increases the moment of the axial 
forces. 

Experiment has proven that when the length of the bar 
does not exceed ten times its least dimension, it is more 
liable to fail by crushing, than by bending and crushing 
combined. 

128 



124 MECHANICS OF MATERIALS 

Therefore, when the length of a bar under axial com- 
pression does not exceed ten times the least dimension, 
the formula P = AS may be used to determine the rela- 
tions between the load and unit stresses. 

Such a bar is called a short column or strut. When the 
length of the bar exceeds this limit, it is called a long 
column; and as the stress which is a combination of bend- 
ing and compressive stresses is not uniformly distributed 
over the area of the cross section, the formula P =AS no 
longer applies. 

The formulas expressing the relations between the loads 
and the dimensions of a long column that are in more or 
less general use are: 

Euler's Formula. This formula is derived from theoret- 
ical assumptions. The magnitude of the load given by its 
use depends on the elastic properties of the material and 
not on the unit stress induced. 

Rankine's Formula. In the derivation of this formula 
certain assumptions more or less theoretically true are 
made. The loads obtained by its use are made to agree 
with the results of experiment by the introduction of an 
empirically determined constant. 

The Parabolic Straight-line Formulas and their modi- 
fications. These formulas are entirely empirical. They 
are simply equations for curves whose coordinates satisfy 
experimental data. 

Art. 69. Euler's Formula. 

The derivation of this formula assumes : that the column 
is perfectly straight, the resistance of the material in any 
cross section uniform, and that the load acts along the 
gravity axis of the column. 



LONG COLUMNS 



125 



Such a column would never bend under any load, but 
would fail by crushing. To produce incipient bending he 
assumed a slight lateral force to be exerted against the 
column while it was under the axial load, and determined 
the value of the axial load P that would keep the column 
bent after the lateral force was removed. 

The elastic curve of a bent column may take any one of 
several forms, depending on the condition of the ends. 



Columns with '' Round " or " Pin " Ends. 

If the ends of the column are de- 
signed so that there will be no restraint 
to the tendency to rotation about a 
point on the end, the column is said to 
have '-'- Round " or "Pm " ends. If the 
ends are rounded as in Fig. 69, it will 
bend in a single curve, and the elastic 
curve may be represented as in the 
figure. 

Taking the origin at (9, the moment 
of P about a point in any section dis- 
tant X from i^ M = Pa — Py, and 
the equation of the elastic curve for 
this case is 




EI^=Pa 
dx^ 



Py. 



(1) 



Multiplying by 2 -^ and integrating with respect to 2/, 



El{^^ = 2 Pay - Pf + C^. 



(2) 



Evidently -^ is when y = 0, hence C. = 0. 
dx ^ 



126 MECHAiNICS OF MATERIALS 

Extracting the square root and transposing, 



dx 



=Vf 



dy 



-P V2 ay — 2/2 



integrating, 



ay-y^ 



X 



= V"^ vers"! ^ + C'g. 



When a^ = 0, then y =^ a and vers"! — = 

a 



(3) 



(4) 



2' 



hence 01, = \/ — • 

2 2 ^ P 



Also when a: = - , y = 0, and vers"! ^ = 0, 



hence L= -TlJ^^ or P = 



2 ^ P 






which gives the value of the load P that will keep a 
column with round ends bent, after the lateral force has 

been removed. As the deflection y has 
disappeared from the expression for P, 
its value is independent of the amount 
of bending. 

Colixmns with Square, Flat, or Fixed 

Ends. 

When the column is so designed 

that the tangent to the elastic curve 

at each end will be parallel to the 

length of the column, the column is 

said to have square, flat, or fixed ends. 

A column of this type may have a cap, 

which on account of the large surface 

it presents, tends to prevent free bend- 

FiG. 69a. ing about a point in that end, or it may 




LONG COLUMNS 127 

have the ends firmly imbedded in masonry. The form 
that the elastic curve takes in this case is similar to that 
of a beam fixed at both ends when there is a single 
load at the middle of the span. Let Fig. 69a represent 
a column with fixed ends as bent by the load P. The 
moments that keep the ends of the column vertical being 
denoted by M^ and M^^ the moment due to M^ and the 
load P, at any section of the column distant x from O, is 

M= M^+Pa-Py, 

and the equation of the elastic curve is 

El'^=M^-\-Pa-Py, (5) 

Integrating and determining the values of Mi and the 
constant of integration from the conditions that dy/dx = 
when y = or a, X = when y = a and x = 1/2 when y = 0, 
we have : 




or P = 



P 



which is Euler's formula for a long column with both 
ends fixed. 

Columns with Round and Square Ends. 

When one end of a column is restrained so that the 
tangent to the elastic curve at that end is always vertical, 
and the other end is left free to rotate about a point in that 
end, the column is said to be one with round and square 
ends. 

The elastic curve will take some such form as shown 
in Fig. 696. 



128 



MECHANICS OF MATERIALS 



Taking the origin at Oi, we have 



EI^^ = Pa- Py + Hx, 



which reduces to 



P = 



2EI 



IT 



P 



(nearly) . 




If the effect of the couple H-H is 
neglected, the value of P is given by 

^ 4: P ' 

which is the value usually given for 
Pwhen the column has one end round 
and the other square. 

In the solution of problems by the 
use of these formulas, I is always the 
moment of inertia about a gravity 
axis perpendicular to the direction of 
bending. If the column is free to 
bend in any direction, the least value 
of I for that section must be used. E is the modulus of 
elasticity, and as the tabulated values are in inch pounds, 
P must be expressed in pounds and I and I in inches. It 
will be noticed that the formulas for the three types of 
long columns differ only by a constant. 

If the strength of a column is defined as the load it will 
carry, and the strength of a column with square ends is 
taken as unity, it is easily shown that the strengths are as: 
1 for a column with square ends, 
^ for a column with round ends, 
-^Q for a column with round and square ends. 



LONG COLUMNS 129 

Ar^ may be substituted for Tin the general formula for 
long columns and the formula written 

A Z2 ' 

where m has values of 1, 4, |^, according to the condition of 
the ends of the columns, and r is the least radius of 
gyration. * 

If the stress was uniformly distributed over the section 

p 

of the column, the value of — would be the unit stress. 
p A 

Taking -— as the unit stress, m = l, tt^ = 10, and the 

, I 

ratio of - = 100, the value of jS for a wrought iron column 

is 25,000 lb. /sq. in., which is practically the elastic limit 
of the material. 

If - is taken as 50, the value of S becomes greater than 

r 

the ultimate strength of the material. 

As the majority of columns in engineering structures 

have the ratios of - from 50 to 150, it is plain that 

Euler's formula will not always give satisfactory results. 
The results of experimental work on long columns show 
that the formula can not be depended on unless the ratio 

of - is nearly 200. 

The ratio of the length to the least value of the radius 
of gyration is called the " slenderness ratio " of a columi 
and when this ratio is greater than 35, the column may be 
considered as a long column. 

* P/A is often referred to as the unit load on the column. 



130 MECHANICS OF MATERIALS 

Euler's formula not being satisfactory for the usual 

range of the values of - has led to the adoption of a 

r 

formula, first proposed by Gordan and later modified by 
Rankine. 

Rankine's modification of Gordan's formula is in com- 
mon use among American engineers, while the European 
engineers prefer Euler's formula or some modification of it. 

If Euler's formula is used for the design of a long col- 
umn, the resulting dimensions should be checked by the 
formula for axial compression to determine the value of 
the unit stress. 

Art. 70. Rankine's Formula. 

This formula is based on tlie assumption that the maxi- 
mum unit stress in any section of a long column under 
axial compression is a combination of the compressive unit 
stress due to the axial compressive force and the maximum 
unit bending stress due to the probable moment of the saine 
force. 

Let Sc be the maximum unit compressive stress in any 

p 

section of a long column, aS = — be the unit compressive 

Mc 

stress due to the load P, Sd = — — the maximum unit com- 
pressive stress due to the probable moment of the axial 
force, and A the area of the section of the column; then 

P Mc 

Sc = S + S,ovSc = ^-\-^^, (1) 

Ji. 1 

from which the maximum unit stress Ss may be found 

P Mc P 

when — and — r- are known. The value of — may easily 
A I A 



LONG COLUMNS 131 

Mc 
be found, but the value of — - is indeterminate, owing to 

the lack of knowledge of the probable eccentricity of the 
force P. 

If we assume that A is the maximum deflection due 
to the unknown moment M, the value of that moment 
may be expressed as PA. Writing Ar^ for / and making 
M = PA, equation (1) reduces to 

By analogy with beams where the maximum deflection 

varies as — , A may be taken as proportional to — . If 
c c 

is some number depending on the material of the column 

for its value, and n is a number whose value depends on 

the conditions at the ends of the column, then A = n4> -. 

c 

Substituting this value of A in equation (2) , we have 

■& = f(l + «0 5),OrP = ^i^, (3) 

1 + n0- 

which is Rankine's formula for the solution of problems 
relating to long columns. Since S^ is usually given in 
pounds per square inch, P must be expressed in pounds, 
and I and r in inches. The value of r to be used will 
depend on the direction in which bending takes place. 
If there is no external restraint to bending, the least 
value of r for the section considered must be used. The 
value of (f) has to be determined by experimental work 
on long columns. The method generally taken to deter- 
mine the value of (j) is to find by experiment the load 



132 MECHANICS OF MATERIALS 

that will cause the column to fail, and substitute that 
value for P in Equation (3). S^ is taken as the ultimate 
compressive strength of the material and, as n^'l^ A, and 
r are all known, the value of ^ can be found. The relia- 
bility of the formula depends on the accuracy with which 
the value of (/> is determined. The following table gives 
the usual value of (j> as found for columns where the ratio 

of - varied from 20 to 200, and therefore can be applied 

^ . Z . . . 

to problems where the ratio of - is within these limits. 

r 



For hard steel 


4> = 


1 

20000 > 


for mild steel 


<^ = 


1 . 

30000 ' 


for wrought iron 


<#> = 


1 
36000 ' 


for cast iron 


4> = 


1 . 
6400 ' 


for timber 


^ = 


1 
3000* 



From Euler's formulas for columns with ends square, 
round, round and square and equal moments of inertia it 
i-s easy to see that the strength of a column with round 
ends is J that of the one with square ends, while the one 
with the ends round and square has but ye ^^e strength of 
the one with square ends. ' 

If P, h and h are the lengths of three columns with equal 
moments of inertia, equal strengths, same materials and 
ends square, round, and round and square respectively, 

smce F - ^2 - ;,2 - 4 1^2 > 

it is evident that I = 2li = ^h. 

AS 
Substituting these values for lmP = — successively, 

l-\-n4>~ 



LONG COLUMNS 133 

there results a numerical coefficient for the last term in the 
denominator of 1, 4, and y-. In Equation (3) n was a 
number depending for its value on the conditions at the 
ends of the columns, hence n must equal 1 for columns 
with square ends, 4 for columns with round ends, and ^^- 
for columns with round and square ends. 

In the use of Rankine's formula the value of So may be 

taken as the safe working unit stress, and then P will be 

P 

the load that can be carried with safety. — is always less 

12 
than aS'c, as 1+^0 — is always greater than unity; hence, 

no matter how short a column may be, Rankine's formula 
will give a safe value for P when the safe value of S\ is 

used. Since nxj) is a small quantity, if the ratio of - is 

small, l-\-n(f)-- may be practically unity and the value of 

P very nearly equal to AS^. The value of (/> being 
derived from experiment, the formula is limited in its 

use to the range in the values of - covered by the exper- 

r 

imental work in its determination. The values of </>, as 
given, apply very well to all columns where the ratio of 

- lies between 20 and 200. Above tliis ratio the load 
r 

given by Rankine's formula will still be safe, as the 
formula gives the value of P too small. 
Art. 71. RittCi's Formula. 

The following formula for the value of the unit load on 
a long column was proposed by Ritter in 1873. 

P ^ S 

'A 1 , j^i!' 



134 MECHANICS OF MATERIALS 

S, the maximum unit stress, is assumed to be equal to or 
less than the elastic limit of the material, *Se the normal 
elastic limit, E the modulus of elasticity, and ix is 47r^, fTr^, 
and TT^ for columns with square, round, and round and 
square ends respectively. 

This formula has the same general form as Rankine's, 

Se 



the empirical constant being replaced by 



fxE^' 



S 
The values of — ^ for materials having a well-defined 
}jlE 

elastic limit, which is approximately one-half the ultimate 

strength, are practically the same as the experimentally 

determined values for for the same materials. For 

materials where the ratio of the ultimate strength to the 

elastic limit is not well defined and greater than 2, such as 

S 
cast iron and timber the values of — ^ differ so widely from 

jjiE 

the empirical values for <f) that that value of results obtained 

by the use of Hitter's formula is very doubtful. 

Ritter's formula may be modified by the introduction of 
a factor in the last term in the denominator, so that the 
results obtained by its use will agree fairly well with the 
results of experiments. 

To determine this factor consider a long column with 
square ends carrying a unit load numerically equal to the 
elastic limit of the material. 

p . 

Writing Se for — in Euler's formula for this column and 
A 

solving for — , we get, — = -—^. From Rankine's formula 

for the column under the same load it is evident that the 

P o g 

value of 1 + 0— must be equal to -^. If g = -^ then 



LONG COLUMNS 



135 



12 , y.2 

q — 1 = </) — . Substituting the value of — as obtained 
from Euler's formula, = \ ^^ ^ — , and using this value 



4^' 



TT 



of </) in Rankine's formula^ we have, 

P ^ Sc 

A 



1 + 



Se{q - 1) l^' 

4&2 ^2 



This formula is similar to Ritter's, and is true when the 
slenderness ratio is such that Euler's formula will give a 
unit load numerically equal to the elastic limit. While 
this slenderness ratio is larger than is ordinarily found in 
practice, the close agreement between the values of as 
determined by experiment and those calculated from 
Se{q - 1) 



</> = 



4^7r2 



may justify the use of the latter expression 



for when the value of that constant has not been experi- 
mentally determined for the conditions obtaining in any 
given problem. 

The following table gives the comparative values of 
Seiq - 1) 



and 



^TT^E 



with the data used in calculating its value. 



Material 


Elastic 

Limit 

Lbs./Sq. In. 


Ultimate 

Strength 

Lbs./Sq. in. 


Modulus of 
Elasticity- 
million 
Lb./Sq. In. 


Se(q-l) 





Cast Iron 

Timber 


4,000 

3,300 

25,000 

30,000 


80,000 
10,000 
55,000 
60,000 


12 
1.5 

28 
30 


1 

6,300 
1 

8,950 
1 


1 

6,400 
1 


Wrought Iron. . . 
Structural steel. . 


3,000 

1 


37,000 
1 


36,000 

1 


40,000 


30,000 



136 MECHANICS OF MATERIALS 

Art. 72. The Parabolic Formula. 

I P 

Writing x for — and y for — , Euler's may be expressed 

as 2/ = — 7r~ ^^^ the curve representing the equation 

plotted using the values of x and y as coordinates. 

P 

If on the same diagram the values of — for varying 

values of -, as determined by experiments on long columns, 
r 

are plotted it would be noticed that for values of - some- 

r 

what greater than 150, Euler's formula satisfies the exper- 

mental data very closely. 

Using the same notation, Rankine's formula becomes 

Sc 

y = 



1 + (^^2 



which as the value of (f)X^ is generally less than unity is 
approximately equal to 

y = Sc — Sc^x'^' 
The last equation may be written 

P P 

y = a — hx^ or — = a — h —, 
A r^ 

which is Professor J. B. Johnson's possible formula for the 

strength of long columns. 

Professor Johnson assumed that for very short columns 

I P 

where th6 ratio of - approached zero, — should be equal to 

r A 

the elastic limit of the material and that the parabola 

representing his formula should be tangent to the curve 



LONG COLUMNS 



137 



given by Euler's formula, and determined value for a and 
b to satisfy these conditions. 

These conditions give a equal to the elastic limit of the 

material and b = - — ^— -. 

In the following table, taken from Johnson's Modern 
Framed Structures, a is the observed elastic limit and the 
value of b is calculated on the assumption that m-K^ =16 
and 25 for hinged and square ends respectively. 



Material 


Lb./Sq. In. 


6 


I = 
7< 


Wrought Iron: 

Hinged ends 

Square ends 

• 

Mild Steel: 

Hinged ends 


34,000 
34,000 

42,000 
42,000 

60,000 
' 60,000 

a 

25,000 
3,300 
4,000 
3,500 


0.67 
0.43 

0.97 
0.62 

6.25 
2.25 

h 

0.6 
0.7 
0.8 
0.8 


170 
210 

150 


Square ends 

Cast Iron: 

Round ends 


190 
70 


Square ends 

Timber (square ends only) : 

White pine • •• • 

Yellow pine, short leaf 

Yellow pine, long leaf 

White* oak 


120 
I 
d 

60 
60 
60 
60 







p p 

The parabolic formula for timber is — = a — b — in 

A d? 

which d is the least dimension of the column. When the 

P 

values of a and b given in the table are used, — is the 

value of the unit load. To determine the safe unit load, 
the values of a and b are to be divided by a factor of safety. 



138 



MECHANICS OF MATERIALS 



Art. 73. Straight Line Formulas. 

Mr. T. H. Johnson after a study of the experimental 
data from tests on long columns found that within the 

usual limits of - the equation 
r 

y = c — ax or — = c — d- 
A r 

could be made to satisfy the experimental data very closely. 
The straight line was assumed to be tangent to the curve 
representing Euler's formula and the values of c and d 
determined so as to satisfy this condition and make the 

p 

values of — agree with the experiments. 

The following table taken from Merriman's Mechanics 
3f Materials gives the values of c and d for different 



Material 


Lb./Sq. In. 


d 


I = 
7< 


Wrought Iron: 

Square ends 


42,000 
42,000 
42,000 

52,500 
52,500 
52,500 

80,000 

80,000 

80,000 

5,400 


128 
157 
203 

179 
220 

284 

438 
537 
693 

28 


218 


Hinged ends 


178 


Round ends 


138 


Structural Steel 




Square ends 


195 


Hinged ends 


159 


Round ends 


123 


Cast Iron: 

Square ends 


122 


Hinged ends 


99 


Round ends 


77 


Oak square ends 


128 







LONG COLUMNS 139 

materials and the limiting value of — for each case. For 

r 

value of - greater than the limiting value, Euler's formula 

may be used to determine the unit load. 

On account of the simphcity of this formula it has been 
used to a considerable extent in certain kinds of work. 
The formula gives the value of the Hmiting unit load for 
the column. To determine the safe unit load the values 
for c and d as given in the table should be divided by the 
factor of safety. 

Many formulas of the same form are made parts of 
specifications and building laws. In the majority of such 
cases the values assigned to c to d are such as will determine 
the safe allowable unit load. 

Art. 74. Comparison of Formula. 

The purely theoretical formula of Euler does not give 
limiting loads that agree well with practice for columns of 
ordinary lengths. When the length of the column is such 
that the strength depends more on the stresses due to 
bending forces than those due to the axial compressive 
forces, the conditions approximate the assumptions made 
by Euler, and his formula will give reliable results. 

Although Rankine's formula is derived from assumptions 
that obtain in every long column, the formula must be con- 
sidered empirical since it contains the experimentally deter- 
mined constant </>. When the experimental work from 
which </) was derived was carefully carried out and the 
conditions under which the experimental columns were 
loaded approximate those for the actual column, the use 
of Rankine's formula leads to very satisfactory results. 

For wrought iron and steel columns having slenderness 



140 MECHANICS OF MATERIALS 

ratios within the limits of Rankine's formula, Ritter's 
modification of that formula gives practically the same 
loads as Rankine's. As the expression which Ritter uses 
to replace depends on the elastic limit and modulus of 
elasticity of the material, it would seem reasonably safe 
to use this expression to determine the value of for 
slenderness ratios in excess of the limiting ratios of the 
experimental work on columns of those materials. 

For the same reasons the additional modification pro- 
posed here, where the value of 4> also depends on the ratio 
of the ultimate strength to the elastic limit of the material, 
should give a wider range to the use of Rankine's 
formula. 

The parabolic and straight line formulas for the limiting 
loads for long columns are entirely empirical and the value 
of the results obtained by their use depends entirely on the 
accuracy with which the straight line or parabola represent- 
ing these formulas satisfies the experimental data. 

In each case the value of the slenderness ratio for which 
the straight line or parabola becomes tangent to Euler's 
curve is the limiting value of the ratio for their use. 
For greater ratios Euler's formula should give reliable 
results. 

The formulas for the strength of beams and bars are 
derived on assumptions that can be closely approximated 
in practice, but the formulas proposed for the unit load 
for a long column give the limiting value of that load under 
conditions as nearly ideal as it is possible to obtain experi- 
mentally. As it is impossible to make any assumption as 
to the variation of the actual conditions from the ideal, 
safety requires the use of a larger factor of safety with 
column formulas than with the formulas for bars and 
beams. 



LONG COLUMNS 141 

In the choice of a factor of safety it must be remembered 
that while with Rankine's formula the factor of safety is 
applied to the ultimate strength of the material, the values 
of the constants in the straight line and parabolic formulas 
are based on a unit stress only slightly larger than the 
elastic limit, and therefore a factor of three for the two 
latter formulas is the equivalent of one of five or six applied 
to Rankine's. 

Cast iron columns are common in engineering work 
on account of the large compressive strength of the 
material. As long as the unit stress due to the bend- 
ing does not equal the unit compressive stress due to 
the load there is no tensile stress in the column. The 
columns are generally made hollow and round in section. 

Wrought iron and steel pipes are also often used for 
columns. The values of / or r may be found as soon 
as the internal and external diameters are known. 

Rolled steel shapes, in channel and I beams, are 
joined together by plates, which are riveted to them, 
and used as columns. Timber is used in the solid 
section and in the hollow box section. In the case of 
circular sections the value of / is the same for bending 
in all directions. When the column is " built " up, as is 
the case when the rolled steel shapes are used with the 
joining plates, the spacing should be such that the 
values of I for the " built " section will be equal about 
the two principal axes. The same is true of the wooden 
box sections. For the rolled steel shapes the value of 
/ for the gravity axes of the rolled section, parallel and 
perpendicular to the web, are given in the tables, and 
must be transferred to parallel axes passing through the 
center of gravity of the built section. For the sections 



142 MECHANICS OF MATERIALS 

where the elements are rectangular the value of the 

o 

gravity moment of inertia is — — , where d is the dimension 
perpendicular to the axis. 

EXAMINATION 

Explain why the formula P = AS does not give results 
that agree with experiment when the bar under axial com- 
pression is very long. 

Define a Long Column. 

What assumptions were made when the formula 

P = m — -— was derived ? 

What is meant by the expression " a column with round 
ends"? "a column with square ends" ? "a column with 
found and square ends " ? 

Derive Euler's formula for long columns with square 
ends ; with round ends. 

Explain why the formula as derived for columns Avith 
round and square ends is approximate. 

Explain why the use of Euler's formula for long 
columns does not always give satisfactory results. 

Give the conditions under which the use of Euler's 
formula for long columns will give satisfactory results. 

State the assumptions that were made for the deriva- 
tion of Rankine's formula for long columns. 

Derive Rankine's formula for the strength of long 
columns. 

State the meaning of each symbol and the units to be 
used in making substitutions in the following formulas : 

P JSr^TT^ , P S^ 

and —r = 



A l^ A ^ l^ 

1 + ^*^ 



LONG COLUMNS 143 

Does Rankine's formula for long columns give results 
that are more reliable than Euler's ? Why ? 

On what does the reliability of Rankine's formula 

depend ? 

Mc 
The formula aS^ = —~ was used in the derivation of 

Rankine's formula for long columns. Does this fact limit 

the use of the formula to materials which satisfy the con- 

Mc 

ditions for the use of the formula aS' = -— ? 

Show by the use of Euler's formulas that the strength 
of a column w^itli square ends being taken as unity, the 
strength of a column of the same size with round ends is 
■J^ and that of a column with round and square ends is -^. 
Explain why the value of n in Rankine's formula is 
1 for a column with square ends 
4 for a column with rounds ends, and 
^ for a column with round and square ends. 
Show that Rankine's formula applied to the solution for 
any column, no matter how short, will always give a safe load 
for that column. Is the same true for Euler's formula ? 

Is there any limit to the length of a column to which 
Rankine's formula will apply? 

Why is the ratio of - used as limiting the use of 
either formula ? 

If the section of a column is a rectangle having one 
side 4 in. and the other 6 in., find the values of I and r 
to be used in the formulas for the strength of long 
columns. 

Why will a hollow cylindrical form make a stronger 
column than a solid cylinder of the same section area ? 

Given the moment of inertia about an axis through the 
center of gravity, how can you find the moment of inertia 
of the section about an axis parallel to the gravity axis ? 



144 MECHANICS OF MATERIALS 



PROBLEMS 

1. A wooden column 10 ft. long is rectangular in sec- 
tion and has round ends. The sides of the rectangle are 
6 and 8 in. respectively. What is the maximum load the 
column will carry? 

Find the safe load. (Use Euler's formula.) 

Solution. The formula is P = -^^, /= M! =1^1'^ M4, 

Z^ 12 12 ' 

since the least moment of inertia must be used. 

E may be taken as 1,500,000 and tt^ used as 10. I = 120 in. 

Substitutina:, 

P = 1.500,000 X 144 X 10 ^ ^ ^^^ 

14,400 

This is the greatest load that can be carried without failure by 

bending. The corresponding unit stress due to the axial force P id 

— = ^ = 3130 Ib./sq. in., or a stress about one half the ultimate 

^48 / H » 

strength of the material. In order to carry this load the column 

must be straight and the load axial. As these conditions are rarely 

ever satisfied, a factor of safety of at least 5 should be applied to the 

result. Using this factor, the safe load is 30,000 lb. and produces an 

axial unit stress of approximately 600 Ib./sq. in., which has a fair 

margin for safety. 

2. If the column in problem l was cast iron and had 
square ends, determine the maximum load it will carry 
and the unit stress induced by that load. Is it possible 
for the column to carry the load ? 

3. Show that a cast iron column with square ends must 

have the ratio of - approximately 78 in order that the 
r 

axial unit stress corresponding to the load P in Euler's 
formula shall be less than the ultimate compressive 
strength of the material. 

4. If the axial unit stress produced by a force equal to 
the value of P as derived by Euler's formula is equal to 

the elastic limit of mild steel, find the ratio of L for the 



' LONG COLUMNS 145 

column. Consider the ends round ; square ; and round 
and square. 

5. A standard 12 in. I beam weighing 35 lb. /ft. is 
to be used as a column with round ends. The length is 
10 ft. What load may be carried with a factor of safety 
of 5? If there should be a factor of 5 used with the 
formula for axial compression, is the load given by Euler's 
formula safe ? 

6. A cast iron column 20 ft. long has a hollow cir- 
cular section. The external diameter is 10 in. and 
the internal diameter is 8 in. Determine the value 
of the maximum load that can be carried, if the allowable 
unit stress for axial compression is 4000 Ib./sq. in. 
What will be the factor of safety against failure by bend- 
ing? (Column has square ends.) 

7. Solve problem 1 by the use of Rankine's formula. 

AS 1 

Solution. The formula isP = ^-r^> ^ = 48 sq. in., <f) = kwtta, 

1 + icfy-L ^^^^ 

Sc = 10,000 lb. / sq. in. for a maximum load, r^=: L = IM = 3 / = 120 

A 48 

in. Therefore -^= 4800, and substituting these values, 



P= 48x10,000 ^65,000 lb. 
1 + 3 o%o 4800 

Comparing the results obtained from the solution by the two 
formulas, we see that the allowable load from the former is about 
twice that obtained by the use of Rankine's formula. Therefore a 
factor of safety of 5 on Rankine's formula is equal to using a factor 
of 10 with Euler's formula. 

8. Find the size of a circular wooden column 12 ft. 
long to carry a load of 50 tons with a unit stress 
of 1200 Ib./sq. in. Column has square and round 
ends. 

9. A wooden column with square ends has a section 
shown in the figure. Find the value of the unit load that 



146 



MECHANICS OF MATERIALS 



may be carried with a unit stress of 
1000 Ib./sq. in. Use Rankine's 
formula. 

10. Find the unit load for the 
column given in problem 9, using 
Johnson's parabolic formula. 

11. Two standard 12-in. chan- 
nel beams are to be used for a 
column. The channels are to 



^3'-^ 



«si 



.^- 



. 1 



^3"-» 



&»/ 



Problem 9. 




Problem 11. 



be placed as shown 
in the figure and 
joine/l by lattice 
work. If the mo- 
ment of inertia of 
the lattice work is 
neglected, find the 
distance between the 
channels so that the 

column will carry the largest load possible. The column 

is to have square ends. 

12. If the safe unit stress in the column given in 
problem 11 is 12,000 Ib./sq. in., find the safe load. 

13. A cast iron column 20 ft. long has a hollow cir- 
cular section. The internal diameter is 8 in. and the 
external diameter is 10 in. Required the load that may 
be carried when the maximum unit stress in the column 
does not exceed 4000 Ib./sq. in. (the column has square 
ends). Compare the result with that obtained from 
problem 6. 

14. The connecting rod for a steam engine has pins at 
either end that are parallel to each other. When bend- 
ing tends to take place in a direction perpendicular to 
the axis of the pins the rod acts as a column with round 
ends. When the bending tends to take place in a plane 



LONG COLUMNS 147 

through the axis of the pins the rod becomes a column 
with square ends. If the section of the rod is rectanguhir, 
find the relation between the depth and breadth, so that 
the column may be equally safe against bending in either 
direction. (Use Euler's formula.) 

15. A cast iron column 20 ft. long has a hollow circular 
section 12 in. outside diameter. The allowable unit stress 
is 10,000 Ib./sq. in. and the load is 50 tons. Required 
the inside diameter of the column. Consider the column 
to have square ends. 

16. A standard 12-in. I beam 40 ft. long has braces 
ilong the web that prevent bending in a direction perpen- 
iicular to the web. If S^ is 12,000 Ib./sq. in., find the 
iafe load for the column. Column has square ends. 

17. Given a timber column 20 ft. long, 8X10 in section. 
The ends of the column are to be considered as square and 
the constants of the material may be taken ultimate 
strength 9000 Ib./sq. in., elastic limit, 3000 Ib./sq. in. and 
modulus of elasticity, 1,500,000 Ib./sq. in. Find the load 
that may be carried when the unit stress does not exceed 
1000 Ib./sq. in. 

(a) By the use of Rankine's formula. 

(6) " " " " Ritter's " and it modifications. 

(c) " " " '' Parabolic '' . 

{d) " '' '' " St. Line " . 



CHAPTER VII 

COMBINED STRESSES 

Abt. 75. Stresses due to Force. 

When a force acts on any material, more than one 
kind of stress may be produced in any fiber of the mate- 
rial. In the previous chapters it was assumed that only 
one kind of stress resulted from the application of the 
force, and the magnitude of the stress calculated on that 
assumption. While this latter condition may be true 
in some cases, the force acting on a bar often produces 
two or more kinds of stress. These stresses when com- 
bined may result in a maximum unit stress greater than 
either of the original unit stresses. Any fiber of a beam 
has a tensile or compressive unit stress due to the bend- 
ing moment, and a unit shearing stress due to the ver- 
tical shear. A shaft carrying a pulley between two 
supporting bearings is a beam and a torsion bar combined. 
Each fiber has a unit stress due to the bending, a unit 
shearing stress due to the vertical shear, and a unit 
shearing stress due to the twisting moment transmitted 
through the shaft. 

Art. 76. Stresses due to Axial Forces Combined with 
Bending Forces. 

When a bar is subjected to the action of forces that 

produce bending only, they induce tensile or compressive 

148 



COMBINED STRESSES 149 

unit stresses that are parallel to the axis of the bar and the 

maximum value of such unit stresses is Sb == — — . If at 

the same time the bar is acted on by axial forces there is 
a tensile or compressive unit stress induced by these 

p 

forces equal to >S = — -. In such cases as approximate 

Jx 

solution of the problem of determining the maximum 
unit stress parallel to the axis is obtained by adding the 
unit stress due to the axial forces and the maximum 
similar unit stress due to the flexural forces. If Sm is the 
maximum unit stress parallel to the axis then Sm = S -\- Sb 

P Mc 

= — - H — , in which *S, Sb and Sm are the same kind of 

stress, is the approximate value of the combined unit 
stress. 

When the axial force is a tensile force, the maximum 
stress found in this way is too large, as the axial force 
on the bent beam produces a moment PA that tends to 
reduce the bending moment of the flexural forces. See 
Fig. 76. When the axial force produces compression, 
the moment PA of the axial force tends to increase the 
bending of the bar, and the approximate solution gives 
the resultant unit stress too small. As the error is due 
to the moment PA, when the deflection is small the 
error is also small. While the engineer as a rule desires 
to be as nearly accurate in his calculations as possible, 
when the approximate formula is simple and errs on the 
side of safety, it is often used in preference to the more 
exact and complicated formula. A bar is often designed 
to resist a combination of compression and bending stress 
by the use of the approximate formula, but when this 



150 



MECHANICS CF MATERIALS 




is done the resulting dimensions should be used in a more 

exact expression and the actual unit stress determined. 
Let Fig. 76 represent a beam under the action of 

flexural and axial forces. Let M be the maximum mo- 
ment of the flexural 
i ^ — — ^'A ^ forces, P an axial 

force which may be 
either tensile or 
compressive, and A 
the maximum de- 
flection of the beam. 
If Ml and *Si are the 
maximum bending 

moment and unit stress due to the sum of the moments 

M and PA, then 

Mic (M±PA)c 




Fig. 76. 



Si = 



(1) 



By analogy with beams under the action of flexural forces 



a 



SP 



only, the value of A may be assumed as A = — — - without 
*^' ^ Ec 

serious error. This value of A substituted in equation (1) 

gives the unit stress due to bending. 

Mc 



Si = 



/3 E 



(2) 



in which the positive sign is to be used when P produces 
a tensile stress and the negative sign when P is a compres- 
sive force. 

The maximum unit stress then becomes, 



Sm — ~T~i~ '^l 

A 



(3) 



COMBINED STRESSES 151 

where Si has the value given in equation (2), Sm tension 
or compression depending on the kind of stress represented 
by>Si. 

The values of a and /3 for various kinds of beams and 
loadings are given in the appendix. There are no values 
that apply strictly to this case as the bending moment is 
increased by the value of PA. Since A is generally a small 
quantity the error made in assuming that the values of a 
and ^ are those found for beams under flexural forces only 
is very small, hence a and (3 will be determined by the kind 
of a beam and the nature of the loading. 

The percentage of the error due to the use of Sj, instead 

PPa 
of ^1, is rb— T- 100. As the amount of the error depends 

directly on the value of P, and inversely on ^, when P 
is small, and JS large, the error will be so small that it 
may be neglected. The error also varies directly as P; 
therefore it is more liable to be serious when I is large. 

For a timber beam 20 ft. long, 6 in. wide, and 12 in. 
deep, carrying a load of 8000 lb. at the middle and at 
the same time a compressive load of 45,000 lb., the use 
of the approximate formula will result in an error of 
approximately 8 per cent. 

For a steel I beam, 10 ft. long, having a moment of 
inertia of 122, a concentrated load of 6000 lb. at the 
middle and an axial compressive force of 20,000 lb., the 
values of S-^ and S^^ differ by less than one per cent, an 
error too small to be taken into account. 

Art. 77. Roof Rafters. 

The common roof rafter is an example of a beam under 
axial and bending forces. As a part of a truss there is some 



152 



MECHANICS OF MATERIALS 



compression in the rafter, and the weight of the roof and the 
probable snow load produce both bending and compression. 
Let Fig. 77 represent a roof rafter, length I in., carry, 
ing a uniform load of w lb. in. The rafter is in equi- 
librium under the horizontal forces H and H, V acting 




Fig. 77. 



vertically at the wall, and the uniform load. Taking mo- 
ments about a point at the foot of the rafter, 



from which 



HI sin <^ = — cos ^, 
H = -- cot 4>. 



The bending moment of the force H about a point in a 
section distant x from the upper end of the rafter is Hx 
sin (]), while that for the uniform loads on the left of the 



wx 



same section is — - cos (j). The unit stress due to the 

bending forces is 

a _ Mc _ (^Hx sin <l> — ^ wx^ cos <^) g 

The compressive force on the same section is H cos </> due to 
the force H and wx sin </>, due to the uniform loads on the 
section. Hence the total compressive stress in the section is 

JP__ (.ZTcos (}) + wx sin c^) 
A" A 



COMBINED STRESSES 153 

Using the approximate formula, the greatest combined 
unit stress parallel to the axis in any section distant x from 
the top of the rafter is : 

P ^ {H cos (f)-\-wx sm (f)) (Hx sin cl) — ^wx^ cos ^)c 

wl 
Substituting — cot for H and proceeding in the usual 

way the maximum value of S is found to be, 

_ SwP cos (f) wl cosec ^ ,w sin (f> tan (^ 
"*" Ihd? 2^d Wb ' 

In any beam carrying vertical loads and supported at 
points not on the same level, the maximum unit stress 
parallel to the axis is a combination of axial and flexural 
stresses. The forces supporting the beam taken with the 
loads form a system of forces in equilibrium. These forces 
may be resolved into components parallel and perpendicular 
to the axis of the beam. The components parallel to the 
axis will produce tension or compression while the per- 
pendicular components will induce flexural stresses, and the 
value of these stresses may be found when the forces 
acting are all known. 

Art. 78. Eccentric Axial Loads. 

When a beam is to resist axial as well as flexural forces, 
it is nearly always possible to make the point of applica- 
tion of the axial force so that its moment about a point in 
the neutral axis of the mid-section of the beam will be 
equal and opposite to the moment of the flexural forces. 
If P (Fig. 78) is an eccentric axial force, and ?/ is the dis- 
tance of its point of 
■^--> application above or 
— Z ~ \^ P below the neutral 

plane, then before 
any bending takes 



<£- 



T T 

I Fig. 78. | 



154 



MECHANICS OF MATERIALS 



place Py is the moment of the axial force about a point in 
the center of the beam. Let M be the maximum moQient 
of the flexural forces ; then if the axial force is compression 
and the distance y measured below the neutral plane has 
such a value that M= Py^ the resultant bending moment 
will be zero. Similarly, when the axial force is tension, 
if 2/ is measured above the neutral surface and its value 
taken so that Py = M^ the resultant bending moment will 
also be zero. 

When the axial force is compressive and the maximum 
deflection due to the flexural forces is greater than the value 
of y determined as above, the value of y should be made 
equal to that maximum deflection. 

Art. 79. Shear and Axial Stress. 

When a bar which is subjected to an axial stress is 
acted on by forces at right angles with the axis, there are 
tensile or compressive and shearing stresses at every point 
in that bar. Let Fig. 79 represent an elementary cube 
cut from any 
portion of the A^V 

bar at which 
there are tensile 
or compressive 
stresses par- 
allel, and shear- 

ino^ stresses ^ _„ 

o Fig. 79. 

perpendicular to the axis. 

The equal tensile or compressive forces Ti, T2 act per- 
pendicular to opposite faces of the cube, while the equal 
shearing forces Vi, V2 act in the same opposite faces and 
are perpendicular to Ti, 7^2. 



H 



r«- 



r 




COMBINED STRESSES 



155 



The cube is not in equilibrium unless a pair of equal 
shearing forces Hi, H2 are introduced. As the cube is in 
equiUbrium and the arms of the couples are equal, it follows 
that V = H. Since the elementary block was a cube, the 
unit stresses due to the equal forces H and V must be equal. 
Hence at every point of the bar there exists a pair of equal 
unit shearing stresses at right angles to each other, in 
addition to the unit tensile or compressive stresses. The 
tensile or compressive and shearing unit stresses that exist 
at every point in the bar combine and create shearing and 
tensile or compressive unit stresses that are greater than 
the original unit stresses. 

Art. 80. Maximum Stresses. 

To determine these maximum stresses let Fig. 80 repre- 
sent an elementary parallelopiped cut from any portion 
of the bar. 

Let its length be dx^ height c??/, width unity, and the 
faces be parallel and perpendicular to the axis of the bar. 

The area on which the 
tensile or compressive 
forces act is c?z/ times 
unity, while that on 
which the shearing forces 
act, is either di/ or dx 
times unity. The forces 
that act on opposite sides 
may be considered to be 
equal since they differ by an infinitesimal quantity. Let 
Sg be the unit shearing stress and S the unit tensile or 
compressive stress. The force that acts perpendicular to 
the dy face is Sd^ and the shearing force in the same 




Fig. 80. 



156 MECHANICS OF MATERIALS 

plane is S^dy. The shearing force in the dx face is S^dx. 
Let dz be the diagonal, <^ the angle between dx and dz^ S„ 
the unit stress perpendicular to dz, and jS^ the unit shear- 
ing stress along dz. Resolving the forces that act on 
either side of dz into components parallel and perpendicu- 
lar to dz we have 

zSidz^ S dy cos (f) -\- S^ dx cos (f) — iS^ dy sin 0, (1) 

S2 dz = Sg dx sin <j> -^ S dy sin </> + ^S^^^ dy cos ^. (2) 

Divide each of these equations by dz^ make -^ = sin <^ and 

— = COS (f>, and equations (1) and (2) reduce to 
dz 

S\ = S sin (f> cos 4*-{- Sg (cos^ ^ — sin^ <^), (3) 

^2= S sin^ ^ + 2 aS'^ sin </> cos </>. (4) 

Writing the equivalents of sin ^ and cos (j) in terms of 

2 ^, we have q 

Si=^ sin 2 (/)+ /S', cos 2 c^, (5) 

AS'2 = f (1-cos 2 ^)+AS'sin2<^. (6) 

By the usual process the value of that makes *Si* a 

maximum and equal to >Sp is given by tan 2 = -^, while 

2 Os 

the value of ^ that makes *S2 * a maximum and equal to &n 

2 *S 
is given by tan 2 0= — ^. 

Substituting these values of 4> in equations (5) and (6), 
we find that the maximum values of >S2 and &\ are 

* The planes in which the maximum stresses Sn and *Sp are induced 
make angles with the axis of the bar that differ by 45 degrees, since 
tan 2 <i)p = — cot 2 i^w 



COMBINED STRESSES 157 



In these expressions S^ and S may be either tension or 
compression. When the positive sign is used before the 
radical, S^ is the same kind of stress as S. When the 
value of SJ^ is obtained by using the negative sign before 
the radical, S^ is compression when S is tension, and vice 
versa. When there is no shearing stress, Sg = and the 
value of Sn is S, while the value of S^ is ^ jS. For Sg = 
the tan 2 (/> = oo or </> = 45°. (/> is the angle between the 
directions of the tensile or compressive unit stresses and 
the plane on which the maximum stress acts, hence the 
maximum S^ makes an angle of 45° or 135° with the direc- 
tion of jS. The angle that jSn makes with S under the 

same conditions is the tan"^ (^ = 0. This latter expression 

P 

shows that S — — will g'ive the maximum unit tensile or 

A 
compressive stress in a bar on which there are axial forces 

only. 

These formulas are general and apply to all combina- 
tions of . tensile and compressive with shearing stresses 
without regard to the nature of the force producing the 

stresses. 

When the unit stress S is induced by axial forces, the 
p 
value of aS' = — -. The stress S may also be due to a bend- 

ing moment, and in that case the value of S is taken from 
S = — -. Sg may be due to a simple shearing force or the 

result of a twisting moment. 

In the former case the value of tS^ to be used is derived 

from Sg— —, and in the latter case jS^ = — ^. When 



158 



MECHANICS OF MATERIALS 



the bar under axial forces is a long colamn, Rankine\s 
formula for long columns must be used to find the value 

of aS'. This formula may be written aS^=— (1 + n(j)-—L 

from which the value of S may be easily found. 



Art. 81. Horizontal Shear in Beams. 

There is a tensile or compressive unit stress in every 
tiber of a beam that is equal to aS' = -zt and at the same 

time a unit shear- 
ing stress resulting 
from the vertical 
shear. When the 
formula V = ASg 
was derived, it was 
assumed that the 
shearing stresses 
were uniformly 
distributed over 
the section of the 
beam. In the pre- 
vious article it was 




Fig. 81. 



shown that there was a pair of shearing stresses at right 
angles to each other. 

Therefore in any beam there is also a liorizontal unit 
shearing that is equal to the vertical unit shearing stress 
at all points of the beam. To deduce an expression for 
the horizontal unit shearing stress at any point of a beam 
imagine a parallelopipedon cut from the top half of any 
beam (Fig. 81). 

Let the length be dx^ the width 5, the distance of the 
lower side from the neutral axis ^/j, and the distance of the 



COMBINED STRESSES 159 

top of the beam from the same axis e. The faces are to 
be taken as parallel and perpendicular to the axis of the 
beam. 

There are compressive forces acting on each end of the 
elementary block which vary in intensity directly as their 
distance from the neutral axis. Let S be the unit com- 
pressive stress at the upper surface, and dA a differential 

area at any distance ?/ from the neutral axis; then — i/dA 

c 

is the force acting at the distance ^ from the neutral axis. 

The sum of the horizontal forces acting on either end of the 

elementary block is the sum of the compressive stresses 

acting on the same area. Calling this sum 11=— \ ydA^ 

and writing — for — , we have H = -- \ yd A. Let the 

I c I^vi 

bending moments at the ends of the block be Mi and 
M2, and the sum of the compressive stresses on the same 
ends be H^ and IT^. Since the ends are separated by 

dx, Mi--M^= dM; then H^- H^ = ^C ydA, For 

I ^y^ 

equilibrium a force equal to H^ — H^ must be introduced, 
and this force must be equal to the sum of the shearing 
stresses on the area dx times h. 

This sum is a horizontal shearing force, and the unit 

stress is — l— ^. Therefore if Si^ is the horizontal unit 

rJ 71//" f*C 

shearing stress, Sj^ = -— I ydA. From the theory 

dx Ih*^yi 

of beams Vdx = dM ov — — = F. Substituting T^for — — 

in the expression for Sf^^ we have /S'^ = — - J ydA as tlie 

value of the unit shearing stress at a distance y^ from 
the neutral axis. 



160 MECHANICS OF MATERIALS 

The parallelopipedon could have been cut from the 
lower half of the beam where the stress is tension and the 
reasoning would be equally true. The formula gives 
the value of the unit shearing stress at a distance y-^ from 
the axis, in a section for which F'is the vertical shear as 
defined in the chapter on beams. The width of the beam 
at a distance y-^ from the axis is 5, Zis the moment of inertia 

of the entire section, while I ydA is the static moment 

of the area of the section lying above the distance y-^ from 
the neutral axis. If e^ is taken as the distance of the 
center of gravity of the area above y-^ from the neutral axis 

ydA — a^e^, and S/i = —- a^c^. 

For a point y^— c from the axis, a^c-^ = 0, and therefore 
jS\ = at the distance e from the neutral axis. When a^ 
is the whole area above the axis, S^^ will be a maximum for 
that section. The greatest value of Sft for the beam will 
be found at the neutral surface in the section where the 
vertical shear is a maximum, since the value of jS/^ varies 
directly with V. Sitnilarly, there will be no horizontal 
unit shearing stress in the sections where V= 0. It has 
been proven that for any point in a section of a beam 
there was a horizontal unit shearing stress that was equal 
to the vertical unit shearing stress at the same point. 
The expression just derived for Sf^ shows that the hori- 
zontal unit shearing stress is^ a variable quantity in any 
section of the beam, therefore the vertical unit shearing 
stress must also be variable. For a rectangular section, 
breadth 6, depth d, the value of iSf^ for any section is 

- -— instead of S, = — , showing: the maximum horizontal 
2bd ' bd ^ 



COMBINED STRESSES 161 

shearing unit stress, and therefore the maximum vertical 
unit shearing stress is 50 per cent greater than the assump- 
tion of uniform distribution of stress would indicate. In 
determining the conditions for the safety of a beam, if the 

unit stress derived from the formula S — -— is a safe 

stress, the beam will in most cases be safely loaded. 
When a beam is short and deep, the horizontal unit 
shearing stress along the neutral surface may exceed the 
safe unit shearing stress. This is especially true of 
timber beams on account of the low value of the ultimate 
shearing strength of timber along the grain. Hence the 
value of the shearing stresses should always be investi- 
gated, as no beam is known to be safely loaded until the 
unit stresses of all kinds have been determined and found 
safe. 

Art. 82. Maximum Stresses in Beams. 

In the general theory of beams as presented in Chap- 
ter III, the' sliearing stress due to the vertical shear was 
assumed to be uniformly distributed over the area of the 
section of the beam. That this assumption was not 
strictly true is evident from the equality of the variable 
value of the horizontal unit shearing stress with the unit 
vertical shearing stress in the same section. The value 
of S/^ being a maximum or zero, as V is a maximum or 
zero, shows that S;^ is zero when ilf" is a maximum, or that 
there are no shearing stresses in the section for which M 

is a maximum. The value of S as derived from S = —— 

will, therefore, be the true unit stress for any section 
where Ji" is a maximum bending moment for the beam. 



162 MECHANICS OF MATERIALS 

The shearing stresses in the fibers along the upper and 
lower sides of the beam are also zero, since a^c^ = 0. 
Hence the unit stresses in these fibers at the various sec- 
tions of the beam will be the value of S ss derived from 

— ^, when M is the bending moment for the section con- 
sidered. 

The unit tensile or compressive stress being zero along 
the neutral surface, the unit stress along that axis is one 
of shear, and its value may be found from the expression 
for S/^, For simple beams M is zero at the supports 
and the unit stress at all points of the section is simply 

With these exceptions the unit stress at all points in 
a beam is a combination of the tensile or compressive 
stresses with the shearing stresses. The value of the 
maximum unit stress at any point in the beam may be 
found from the expressions for the maximum values of 
Sj^ or Sp^ when jSf^ is substituted for S^. These maximum 
stresses make angles with the axis of the beam that 
depend for their value on the relative values of S and Sfi. 

The unit stress given by -— - is the true unit stress, when 

M is s. maximum moment, and it is easy to see by an in- 
spection of the expressions for the maximum values of 
iSji and Sp that their values can rarely ever exceed those 

Mo V C^ 

given by -— - and — I ydA, When a beam is deep 

vertically and carries a concentrated load at the center, 
the value of V is constant for all sections up to the 
middle of the beam. Therefore it is possible in such a 
beam that the maximum value of S^ may exceed the 



COMBINED STRESSES 163 

Mc . . 

maximum value of S as found from . This is espe- 
cially true of beams having I sections, as the value of the 
static moment of the flanges is nearly as large as the 
moment of the whole area above the neutral axis. The 
value of S^ in the web just below the flange being very 
large, for a section just to the left of the load where F^is 
a maximum and S nearly so, the values of S.^ and aS^^ may 
be greater than the maximum values of S and Sj^. 

The shearing stress S^ at any point of the beam is a 
stress that is inclined at angles that vary from 0° to 45° 
with the axis of the beam. When the beam has an I 
section and is deep, these shearing stresses have a tend- 
ency to cause the web to buckle. To resist this tendency 
vertical angle irons are often riveted to the webs. When 
a beam of an I section is composed of angle irons riveted 
to a web plate, making what is known as a plate girder, 
the force on the rivets joining the angles to the web can 
be found from the values of S^ and /Sy^, and the areas over 
which these stresses are distributed. 

While the theory of beams as presented in Chapter III 

was defective in its assumptions regarding the distribution 

of the shearing stress, and its neglect of the combined 

stress, this discussion shows that for the majority of beams 

Me 
the formula S= -— ^iH gjye the maximum bending stress 

when the value of itf is a maximum. The formula may 
be used for the design of all beams and the dimensions 
checked for safety by determining the values of /S'^, /S^, 
and Sf^, 



164 MECHANICS OF MATERIALS 

EXAMINATION 

Give some examples of bars where the forces acting 
produce more than one kind of stress. 

When a bar under bending forces also has a tensile or 
compressive axial force acting on it, why does not the 
resulting unit stress always equal the sum of the unit 
stresses due to each force ? 

Develop the expression 

^ Me 

'^1 ~ ;u^' 



i± 



p E 



where S-^ is the maximum unit stress due to both the axial 
and bending forces acting on any bar. 

What assumptions are made in the development of the 
formula that are not strictly true ? 

When a beam which carries bending loads also has to 
resist tensile or compressive loads, how can the resulting 
bending moment be made practically zero ? 

When a bar is subjected to both axial and shearing 
forces, show that at every point of the bar there is a pair 
of equal unit shearing stresses whose directions make 
right angles with each other. 

What is the effect of the combination of shearing with 
unit tensile stresses ? Is the effect any different if the 
shearing stress is combined with an equal stress in com- 
pression ? 

Show how to determine the value of the maximum unit 
stress when tensile or compressive stresses are combined 
with shearing stress ? 

A bar is being acted on by tensile or compressive forces 



COMBINED STRESSES 165 

applied in the line of the axis. Is there any shearing 
stress ? How may its value be determined ? 

If the maximum tensile or compressive unit stress is 
given by 

and the positive sign is used before the radical, what kind 
of stress is jS^? 

What is meant by the expression " horizontal shear " in 
beams ? 

Deduce an expression for the horizontal unit shearing 
stress at any point of a beam. 

Under what conditions will the unit stress given by 
S=—- be the maximum unit tensile or compressive stress 

in a beam ? 

Under what conditions may the horizontal shearing 
stress become the most dangerous stress in the beam ? 

A deep I beam is short and carries a load at the middle. 
Is it possible that at some point in the beam there may 
be a unit tensile or compressive stress greater than that 

Mc 
given by aS'^ — - when the value of Mis that of the maxi- 
mum bending moment for the beam ? 

When a beam is rectangular in section, where will the 
maximum unit shearing stress be found? Is the same 
true for beams of other sections ? 

PROBLEMS 

1. A roof with two equal rafters has a span of 40 
ft. and a rise of 15 ft. If the weight of the rafter is 
neglected, determine the size of the rafters 6 in. wide 



166 MECHANICS OF MATERIALS 

when each rafter carries a uniform load of 50 lb. per 
foot. 

Assume the allowable unit stress as 600 lb. / sq. in. 

2. Assume the rafters in problem 1 to carry a load of 
1000 lb. at the middle, instead of the uniform load, and 
find the depth of the rafters for the same unit stress. 

3. A simple wooden beam, 30 ft. long, 12 in. deep, and 
4 in. wide, carries a single load of 320 lb. at the middle 
and an axial compressive load of 14,400 lb. Find the 
maximum unit stress in the beam, 

(^) by the approximate method, 
(5) by the more exact method. 

4. A simple steel I beam 20 ft. long, 12 in. deep, 
weighing 35 lb. / ft., carries a uniformly distributed load 
of 500 lb. /ft. and sustains a compressive load of 40,000 
lb. Is it safe if a factor of safety of 5 is needed ? 

5. Find the points of application of the compressive 
loads in problem 4 so that the unit stress resulting from 
the effect of the loads will be as small as possible. 

6. A steel eye bar 30 ft. long has a section area of 
2 by 6 in. The eye bar is to be used horizontally, and 
carries a tensile load of 288,000 lb. Determine the dis- 
tance from the center of the eyes to the axis of the bar so 
that the resulting unit stress will be as small as possible. 

7. A horizontal steel shaft, 5 in. in diameter, is 20 
ft. between bearings, and carries a load of 1200 lb. at 
the center. It transmits 250 H.P. at 100 revs. /min. 
Required the maximum unit stress induced. 

8. ' A standard 2-in. steel bolt is screwed so as to 
cause a unit tensile stress of 10,000 lb. /sq. in. What 
shearing force may be resisted if the maximum unit 
stress is not to exceed 15,000 lb. /sq. in.? 



COMBINED STRESSES 



167 



9. How many 1-in. standard steel bolts must be used 
to resist a shearing force of 300,000 lb. if the tensile stress 
due to screwing up is assumed to be 10,000 lb. /sq. in. 
and the allowable maximum unit stress is taken as 12,000 
lb. /sq. in.? 

10. An angle iron is bolted to the side of an I beam 
and supports another beam whose reaction at the sup- 
ported end is 50,000 lb. Select a number of 1-in. bolts 
to carry the load. Assume that the unit stress in each 
bolt due to screwing up is 12,000 Ib./sq. in., and the 
factor of safety" is 4. 

11. Given an I-beam having flanges and web whose 
cross-sections are rectangles of the dimensions given in 
Fig. 82, used as a simple beam 
20 ft. long and carrying a con- 
centrated load of 30,000 lb. at 
the middle of the span. Find 
the maximum horizontal unit 
shearing stress and the horizontal 
unit shearing &t ess at a. 

12. Given the same beam as in 

problem 11, carrying the same 

loads, find the difference between 

the maximum combined compressive unit stress in a section 

taken just to the left of the central load and the value of 

Mc ^ ^ 

-J- at the dangerous section. 

13. Given the same beam as in problem (11) and carrying 
the same loads, find the total horizontal shearing force 
acting along the neutral surface and to the left of the load. 
What is the value of this shearing force per Hneal foot of 
the beam? 



\( 


/a 


— — -)( 


1 


1 
1 






1 

1 

1 

1 


.ii% 




1 

1 










1 



Fig. 82. 



168 MECHANICS OF MATERIALS 

14. Given a beam having the same section as in problem 
(11) to be used as a simple beam on a span of 8 ft. and 
supporting a load of 160,000 lb. concentrated at the middle. 
Find the magnitude of the maximum combined unit stresses 
in tension and shear. What angle will the maximum com- 
bined stresses make with the axis of the bar? What will 
be the value of the greatest horizontal unit stresses in 
tension and shear? 



CHAPTER VIII 



V 



COMPOUND BARS AND BEAMS 

Art. 83. Definition. 

When a bar is composed of more than one kind of 
material it is sometimes termed a compound 
bar. The formulas derived in the previous 
chapters apply only to bars made of one 
material throughout. This chapter will be 
devoted to the investigation of a few of the 
simpler cases of stress in the compound bars. 

Art. 84. Compound Columns; alternate layers. 

A column or pier built with alternate layers 
of different materials, as in Fig. 84, will evi- 
dently carry only the load the weaker section 
will support. The unit stress in any section 
may be found when the area of the section is known, as 
each section has to support the entire load. 

When the modulus of elasticity of the material and the 
length of each section are known, then the amount of 
shortening for each section can be found as for simple 
bars. The total shortening of the entire column is the 
summation of the deformations of the different sections. 

Art. 85. Compound Columns ; longitudinal layers. 

When a column is composed of different materials 
arranged longitudinally, the column becomes a bundle 

169 



Fig. 84. 



170 



MECHANICS OF MATERIALS 



of simple bars. Each bar does not carry a part of the total 
load that is proportional to its area as in the previous case. 
All the bars have the same amount of deformation, and 
when the unit stress is within the elastic limit, the unit stress 
in each bar must vary directly as the modulus of elasticity. 
Let Fig. 85 represent a section of a hollow steel column 
encased in concrete carrying a compressive load which 
produces axial stress only. Considering 
each part of the compound column as 
a separate bar, the kind of material in 
that part and the necessary factor of 
safety determines the safe unit stress Sc 
This stress depends solely on the factor 
of safety and the material. 
When the two columns act together 
and support the load as one column, the condition that 
each of the columns of the different materials must sustain 
the same deformation, gives the relation 




e = 



flh /2^2 



which may be written 



e = 



El 

Pih 
AiEi 



E, 



P2I 



2^2 



A2E2 



(1) 



(2) 



In equations (1) and (2) /i and /2 are the unit stresses that 
actually occur in the different materials composing the 
column; P, I, A, e, and E have their usual meanings, and 
the subscripts 1 and 2 refer to steel and concrete respect- 
ively. 

When the lengths of the two columns are the same, 
equation (1) reduces to 

fl f2 fi El 

11- = Jji or — = — ^. 

El E2 /2 E2 



e = 



(3) 



COMPOUND BARS AND BEAMS 171 

That is, the actual unit stresses induced in the different 
materials composing the column are in the same ratio as 
their respective moduli of elasticity. As the ratio of the 

safe unit stress — in any two materials is independent of 

02 

the ratio of their moduli of elasticity, it is evident that the 
ratio of the safe unit stresses for the materials will not 
necessarily be equal to the ratio of the actual unit stresses 
induced. Si and S2 are the limiting stresses that can not 
be exceeded with safety. 

In general, one of the safe unit stresses may be used for 
either /i or /2 provided the relation imposed by equation (3) 
gives a value for the other actual stress that is either equal 
to or less than the safe unit stress for that material. 

When the values of /i and /2 have been determined and 
the areas of the different parts known, the partial loads 
carried by each part is determined by Pi = Aifi and 
P2 = A2/2, the total load being given by P,= Pi + P2. 

When only the total load and the materials are known, 
the problem of finding the proper areas for the two columns 
is indeterminate, as any values can be assigned to Pi and P2 
as long as their sum is equal to P. The same reasoning 
may be applied to columns of three or more materials by 
writing equation (1) as 

= /lil = /2^2 _ fsh fJn ,-v 

El E2 Es Eft 

and equation (2) as 

P = Pi + P2 + P3 . . . + P« (6) 

Art. 86. Compound Beams. 

When two or more beams of different materials are 
fastened together so that they act as one beam, the com- 
bination is called a compound beam. 



172 



MECHANICS OF MATERIALS 



Beams having cross-sections similar to those shown in 
Figs. 86a and 865 are examples of compound beams com- 
posed of two materials, wood and steel. The form of the 
section is immaterial as long as the different parts are con- 
nected so as to act as one beam. 

If the connecting bolts are removed, the safe unit stresses 
in each of the component parts of the compound beam 





Fig. 86 a. 



Fig. 86&. 



may be assigned at once, as in this case each beam acts 
independently and the safe unit stresses will depend on 
the material and the factor of safety. 

When the two parts of the beam are bolted together as 
shown in the figures, each part of the beam deflects equally 
and this condition gives the relation 



A = 



OL fih 



a 



f2l2^ 



(1) 



(3 Eici (3 E2C2 
Which when expressed in terms of the partial loads becomes 



A = 



1 Wlh^ 1 W2l2^ 



(2) 



i8 EJi /3 E2I2 

where /i and /2 are the actual unit stresses in tension or com- 
pression induced in the different parts of the beam by the 



COMPOUND BARS AND BEAMS 173 

load carried, and the other symbols have their usual 
meanings. 

Equations (1) and (2) may be expressed as 

and 

Wi EJih^ 



W2 E2l2ll^ 



(4) 



giving the ratio of the actual unit stresses and the ratio 
of the partial loads for each part of the beam. As was 
found to be the case when considering compound columns, 
the safe unit stress for each part of the compound beam 
will depend on the material and the factor of safety. These 
unit stresses are the limiting values for the unit stress for 
each material. They can not be exceeded with safety. 

The value of /i or /2 may t>e assumed as equal to the 
limiting value of the unit stress for that material provided 
the relation imposed by equation (3) does not cause the 
actual unit stress in the other material to exceed the safe 
limiting value of the unit stress for that material. 

When /i and /2 have been determined, the partial load 
carried by each of the component parts of the beam may 

fl 
be found from the formula M = — which in this case may 

be expressed as 

TFi=^ or W2 = i^. (5) 

llCi I2C2 

The sum of the partial loads is the total load on the com- 
pound beam or 

W = Wi -\-W2 (6) 



174: MECHANICS OF MATERIALS 

When the load to be carried by a given compound beam 
is known, the partial loads for the different parts may be 
found by the use of the equations (4) and (6) and then the 
actual unit stresses may be determined from equation (5). 

The problem of designing a compound beam to cover 

a given span and carry a given load is indeterminate. The 

values of the safe unit stress for each material will be 

known, but the relative values for / cannot be found until 

the values for I, c, and I for each of the component parts 

of the beam are known. 

c 
The value of the ratio — for each beam may be chosen 

so that the value of / for each of the component parts will 
be equal to the limiting stress for the material, and if at 

the same time the values of — are chosen so that the equation 

c 

(6) will be satisfied, the design will be satisfactory so far 

as unit stress is concerned, as the unit stress in each part 

will equal the limiting value. 

Writing equations (3), (4) and (6) as 

EiCi E2C2 EsC3 ' ' ' EnCn 

_ Wih^ _ W2h^ _ W.ih^ Wnlr? ,^x 

~/3^l7l ~ ^E2l2 ~ ^Esls ' ' ' ^Enln 

W = W1 + W2 + WS . . . -\-Wn (9) 

it is evident that this discussion will apply equally well 
to compound beams with any number of different materials. 

EXAMINATION 

What is a compound bar? 

When a compound bar is used as a short column, will 
the formula for axial compression always give the true 



. COMPOUND BARS AND BEAMS 175 

maximum unit stress? Explain your answer fully, giving 
reasons. 

When a be^m is composed of more than one kind of 
material, name the conditions that are used to determine 
the part of the load carried by each material. 

Is it always possible to use the safe unit stresses as 
determined by experiments on the materials that compose 
a compound bar or beam to determine the loads that may 
be carried? Give reasons for your answer. 

When the depth of the different parts of a compound 
beam are not the same, will the unit stresses induced be 
in the same ratio as the moduli of elasticity? 

PROBLEMS 

In the following problems the values of the moduli of 
elasticity to be used are: Steel, 30,000,000 Ib./sq. in.; 
wood, 1,500,000 Ib./sq. in.; cast iron, 12,000,000 Ib./sq. in.; 
and concrete, 2,400,000 Ib./sq. in. 

1. A hollow circular cast-iron column 6 in. outside 
diameter, 4 in. internal diameter, was designed to carry 
a given load using 10,000 Ib./sq. in. as the allowable unit 
stress. If the column were enclosed in a square box section 
wooden column 10 in. outside and 6 in. inside, how much 
more load could be carried on the compound column than 
on the cast iron one? Use 1000 Ib./sq. in. as the safe 
allowable unit stress in the wood. 

2. A load of 138,270 lb. is to be carried on a compound 
column formed by filling the center of a circular cast-iron 
column 6 in. internal diameter with concrete. Assuming 
the limiting unit stresses in the cast iron and concrete to 
be 10,000 Ib./sq. in. and 1000/sq. in. respectively, find the 
necessary external diameter of the column. 



176 MECHANICS OF MATERIALS - 

3. Given a circular cast-iron column, 8 in. external and 
6 in. internal diameter filled with wood. Assuming that 
the limiting unit stresses in the wood and cast iron are 
1000 Ib./sq. in. and 10,000 Ib./sq. in. respectively, find the 
load that the compound column will carry. 

4. A steel pipe 6 in. external diameter and a sectional 
area of 5 sq. in., is to be encased in concrete 2 in. thick 
for fire protection. Find the load that may be carried on 
the steel pipe, assuming that the concrete does not carry 
any of the load, and compare the result with the load that 
may be carried when the concrete carries its share of the 
load. Use 15,000 Ib./sq. in. and 600 Ib./sq. in. as the 
limiting unit stresses in the steel and concrete. 

5. A compound beam 20 ft. long is formed by bolting 
two standard 12-in. channel beams to the sides of a timber 
beam 4 in. wide and 12 in. deep. Assume that the safe 
unit stresses for the wood and the steel are 1000 Ib./sq. in. 
and 15,000 Ib./sq. in. respectively and find the uniform 
load that may be carried, neglecting the weight of the beam. 

6. A timber beam 24 ft. long, 10 in. wide and 12 in. deep . 
is to be reinforced by bolting a steel plate 10 in. wide by 
J in. thick to each side of the beam. Find the uniform 
load that may be carried. Assume the hmiting unit 
stresses in the steel and timber to be 15,000 Ib./sq. in. 
and 600 Ib./sq. in. respectively. 

7. A timber beam 20 ft. long, 10 in. wide and 12 in. deep 
is to be reinforced by bolting steel plates 12 in. wide to each 
side of the beam in order that a load of 14,400 lb. at the 
middle of the beam may be carried. Assume that the 
limiting unit stresses in the steel and the timber are 15,000 
Ib./sq. in. and 600 Ib./sq. in. respectively and find the 
thickness of the steel plates. 



, COMPOUND BARS AND BEAMS 177 

8. A wooden beam 20 ft. long, 10 in. wide and 12 in. deep 
has steel plates 10 in. wide and J in. thick fastened to the 
top of the bottom of the beam. Assume that the limiting 
unit stresses in the steel and the wood are 15,000 Ib./sq. in. 
and 600 Ib./sq. in. respectively and that there is no rela- 
tive motion between the steel and the wood. Find the 
uniform load that may be carried, neglecting the weight of 
the beam. 

9. If the uniform load was 60,000 lb., find the total force 
exerted over one-half of the length of the beam that would 
tend to shear the bolts used to fasten the plates. 



CHAPTER IX 
REINFORCED CONCRETE 

Art. 87. Steel and Concrete in Combination. 

Assuming the safe unit compressive stress in steel and 
concrete to be in the ratio of 30 : 1 and the costs of the two 
materials per unit of volume in the ratio of 60 : 1 it follows 
that the cost of the steel. member to support a given com- 
pressive load will be about twice that for a concrete 
member. As concrete is approximately ten times as strong 
in compression as in tension, it is evident that the cost of 
a concrete member to carry a given tensile load will be 
about five times that for a steel member to carry the same 
load. Hence, concrete is the cheaper material when only 
compressive stresses are induced, and steel when the 
stresses are both tensile and compressive. 

When a combination of steel and concrete is used to 
resist tensile forces the full strength of the steel is not 
available unless the unit stress in the concrete is dis- 
regarded; since before the unit stress in the steel reaches 
a safe allowable value, the unit stress in the concrete will 
have reached its ultimate value and the entire load will b3 
carried by the steel. 

Steel beams in bridges and buildings are often encased 
in concrete on account of its fireproofing and non-corrosive 
properties. In the calculations for the strength of such 
encased beams no account is taken of the resistance of 

the concrete. 

178 



REINFORCED CONCRETE 179 

Art. 88. Reinforced Concrete Beams. 

When a concrete beam has steel bars embedded in the 
concrete on the tension side of the neutral surface, the 
beam is called a reinforced concrete beam. 

The same kinds of stresses are induced in reinforced con- 
crete beams as in beams of one material and in addition 
there is a bonding or shearing stress between the steel and 
the concrete. The relations between the loads, stresses 
of all kinds, and the dimensions of a reinforced concrete 
beam must be determined before the design or investigation 
as to the safety of a reinforced concrete beam can be under- 
taken. 

Art. 89. Tensile and Compressive Stresses Parallel to the 
Neutral Surface. 

The common theory for reinforced concrete beams 
assumes that the concrete above the neutral surface carries 
all the compressive stresses and that these stresses vary 
in intensity directly as their distance from the neutral 
surface. On the tension side of the neutral surface the 
tensile resistance of the concrete is neglected, and it is 
assumed that the steel reinforcement carries all the 
tensile stress, the concrete being used to keep the steel 
in its proper position and to resist the horizontal shearing 
stresses. The assumption that there is no tensile resist- 
ance in the concrete is undoubtedly erroneous, but it errs 
on the side of safety. 

The relation between the moment of resistance in a 
section of a beam for which the bending moment is a maxi- 
mum, and the tensile or compressive unit stresses parallel 
to the neutral surface is derived on the following assump- 
tions : 

(1) That the concrete carries all the compressive stresses 



180 



MECHANICS OF MATERIALS 



and that these stresses vary in intensity directly as their 
distance from the neutral surface. 

(2) That the steel reinforcement carries all the tensile 
stress. 

(3) That the unit stress in the steel is constant and 
depends on the distance of its center of gravity from the 
neutral axis. 

(4) That the neutral surface is so located that the tensile 
resistance of the steel is numerically equal to the compres- 
sive resistance of the concrete. 




Fig. 89. 



(5) That any plane section of the beam remains a plane 
during bending. 

(6) That there is perfect adhesion between the steel and 
the concrete. 

Let Fig. 89 represent a portion of a reinforced concrete 
beam, and the section X-X the dangerous section. If the 
value of the resisting moment for this section can be 
expressed in terms of the maximum unit stresses in tension 



REINFORCED CONCRETE 181 

or compression, and the dimensions of the beam, the 
relation between the load on the beam and the unit stresses 
becomes determinate, as the expression for the maximum 
bending moment can always be written and equated to 
the expression for the moment of resistance. 
Let 
a = area of section of steel reinforcement in square 

inches. 
A = hd = area of section of concrete in square inches. 
d = the distance from the outer fiber in compres-. 
sion in inches to the center of gravity of the 
steel. 
h = the breadth of the beam in inches. 
I = length of the beam in inches. 
Es = the modulus of elasticity for steel in pounds per 

square inch. 
Ec = the modulus of elasticity for concrete in pounds 

per square inch. 
Mr = the maximum moment of resistance for the beam 

. in inch pounds. 
M = the maximum bending moment for the beam in 

inch pounds. 
fs = the maximum unit tensile stress in the steel in 

pounds per square inch. 
fc= the maximum unit compressive stress in the outer 
fiber of the concrete in pounds per square inch. 
n = the ratio Es/Eq. 
r = the ratio fs/fc- 

p = the ratio of the sectional area of the steel reinforce- 
ment to the area of the concrete above the center 
of gravity of the steel; p = ahd. 
k = the ratio of the distance of the neutral axis of the 
section to the distance of the center of gravity 



182 MECHANICS OF MATERIALS 

of the steel, both being measured from the outer 
fiber in compression. 
j = the ratio of the distance between the resultants 
of the stresses in the concrete and steel to the 
depth of the beam. 
For equilibrium the resultants of the stresses in the steel 
and concrete must be numerically equal and the moment 
formed by these resultants must be equal to the maximum 
bending moment for the beam. 

Since all the unit stresses considered are within the 
elastic limit, under the asumptions made it follows that 

k 

E. (1 - k)d tih^l^lJl n^ 

h kd JcE, h ^' 

Ec 

f w 

Substituting r for j, n for — ^ and solving for k gives 

k = -^. (2) 

1+- 

n 

The area over which the unit compressive stresses are 
distributed is hkd, while the constant tensile unit stress 
in the steel is distributed over an area phd, and since the 
total compressive force must be equal to the total tensile 
force, 

pdhfs = kdh{-' or r =^f = A_. (3) 

/ ^ Jc ^P 

Substituting the value of k from equation (2) in equation 
(3) and solving for p, 

P=—T^ V- (4) 



2r 



('+y 



REINFORCED CONCRETE 183 

Equation (4) shows that there is a fixed relation 
between the ratios p and r. When either ratio is given 
the value of the other is fixed. 

In the design of a reinforced concrete beam, the steel 
ratio p may be chosen so that equation (4) will be satisfied 
when fs and fc are respectively to the safe unit stresses for 
the two materials. The safe unit stress in compression 
for a concrete of average strength lies between 600 and 
700 Ib./sq. in., while the allowable stress in steel may be 
anywhere between 16,000 and 30,000 Ib./sq. in. Con- 
sidering horizontal stresses in tension and compression only, 
an ideal value of p is obtained when fs and fc are assumed 
as equal to the limiting unit stresses for the steel and con- 
crete respectively. As an increase in the unit stress in the 
steel increases the bonding stress, the actual unit stress in 
the steel is generally limited to 16,000 Ib./sq. in., regard- 
less of the possibility of the steel being able to carry a 
higher unit stress with safety. 

When p has any value other than the ideal one, the 
ratio of fs to fc can not be equal to the ratio of the safe unit 
stresses for steel and concrete. 

Substituting the value of r from equation (2) in equation 
(4) and solving the resulting quadratic for k, 

/b = pnL/i+A_i (5) 

L \ pn J 

When the problem requires the determination of the safe 
load for an existing beam, equation (5) will determine the 
value of k, as the ratios p and n will be given by the data 
for the problem. 

This value for k, substituted in equation (2), will deter- 

fs 

mine the ratio ^ = t- When the value of this ratio is 

Jc 



184 MECHANICS OF MATERIALS 

known and the safe limiting unit stresses are given, the 
maximum values for/s and fc may easily be determined. 

The resisting moment for the section being equal to the 
moment of the couple formed by the resultants of the 
tensile and compressive stresses, jd times either resultant 
gives the value of the resisting moment. 

Hence the value of the bending moment at the dangerous 
section is 

M = Mr =fs V^djd = ^ khdjd, (6) 

in which jd = d ll 1 . 

This theory has been developed on the assumption that 
M was the bending moment for the dangerous section of 
the beam, but it applies equally well to any section when 
M is taken as the bending moment for that section. 

Substituting for M in equation (6), the value of the 
maximum bending moment for the beam and for either 
fs or fc the maximum possible value, will give the relation 
between the loads on the beam and the maximum unit 
stresses parallel to the neutral axis that may be used to 
investigate the safety of any given reinforced concrete 
beam. 

From equation (6) we may obtain a value of hd^ as 

To design a reinforced concrete beam to resist a given 
maximum bending moment and considering horizontal 
stress in tension and compression only, the values oi fs and 
fc may be taken as the maximum unit stress allowable in 
each of these materials when used in reinforced concrete 
beams. 



REINFORCED CONCRETE 185 

The values oi p, j and k may then be determined by the 
use of equations (4) , either (2) or (3) , and j = 




Substituting the proper values for p, j and fs, or k, j and 
fc in equation (7) the value of bd'^ may be determined. 

While it would appear that any values that would satisfy 
equation (7) might be used for b and d, other conditions 
limit the least value of b to ^d. 

Art. 90. Reinforced Concrete T-Beams. 

Since the concrete below the neutral surface is not con- 
sidered as carrying any kind of the horizontal tensile 
stress, any more concrete below the surface than that 
required to resist the horizontal shearing stresses and hold 
the steel reinforcement in position adds unnecessary weight 
to the beam. For this reason beams of the form shown in 
Figs. 90, called T-beams, are in common use. 

In the derivations of the formulas for T-beam sections 
the same notation applies as was used for beam of rec- 
tangular sections with the following exceptions and addi- 
tions : 

b = the width of the flange. 

6' = width of the web. 
t = the thickness of the flange. 

z = the distance of the resultant compressive force 
below the top of the flange. 

p = the steel ratio a/bd and not a -^ area of concrete. 

The distance of the neutral axis below the top of the 
flange will depend on the relative depths of the flange and 
web and on the amount of steel reinforcement. 

As this distance may be more or less than the thickness 
of the flange, the neutral axis may lie either in the flange 
or web. When the neutral axis lies in the flange (Fig. 90a), 



186 



MECHANICS OF MATERIALS 



all the theory used with rectangular sections applies 
equally well for T sections, as in either case any effect 
the concrete below the neutral axis may have on the 
strength of the beam is neglected. 

When the axis lies in the web the conditions are changed 
(Fig. 906). While the unit stress in the concrete is still 
assumed to vary directly as the distance from that axis, 
the position of the resultant compressive stress is not the 
same as for a rectangular section. 

The unit compressive stresses in the web will be small, 
as they are near to the neutral axis, and since the width 
of the flange is generally much greater than the width of 
the web, the compressive resistance of the concrete in the 



I- 




.b'. 



d 

1 



A/eutralAxis 



&OB- 




FiG. 90a. 



Fig. 906. 



web will be small compared with the resistance of the con- 
crete in the flange. 

In the great majority of T-section beams the proportion 
of the compressive stresses carried in the web is so small 
that its neglect will not seriously affect the result, and the 
knowledge that any error made will be on the side of safety 
has led to the development of an expression for the resist- 
ing moment for a T-section reinforced concrete beam, in 



REINFORCED CONCRETE 187 

which the effect of any compressive stress there may be 
in the web is neglected. 

Following the same line of argument as was used in the 
beams of rectangular section the value of k is found to be 

k = ^— . (1) 

r 

1 + - 
n 

The average unit compressive stress in the flange is one- 
half of the sum of the unit compressive stresses at the top 
and bottom of the flange. The unit stress at the top of the 

flange is fc and that at the bottom is /<.( 1 — — j . The 
average unit stress, one-half the sum of these stresses, is 

This average unit stress is distributed over an area equal 
to ht, and since the sum of the compressive stresses must 
be numerically equal to the tensile resistance of the steel, 

f^Pbd=f{t-^^bt. (3) 

This may be written as 

Eliminating r between equations (1) and (4) and solving 
for k gives 

k = ^. (5) 

pn + - 

The resultant of the compressive stresses in the flange 
passes through the centroid of the part of the shaded area 



188 



MECHANICS OF MATERIALS 



lying in the flange at a distance from the top (Fig. 90) of 
the flange equal to 



a 

2/c--, 
a 



(6) 



From Fig. 905 the arm of the couple formed by the 
resultants of the tensile and compressive stresses is 

jd = d - z. (7) 

Using the expression for the arm of the couple and the 
magnitude of the resultant forces from equation (3) the 
moment of resistance is given by 



Mr =fsVhdjd 



f'^'-irr^' 



(8) 



in which the values of /s and fc are such that equations (1) 
and (5) are satisfied. 

Art. 91. Shearing Stresses. 

Let Fig. 91 represent a differential length, dl, of a beam, 
C and C' the resultant compressive forces, and T and T' 

the resultant tensile forces acting on 
each end of the section. 

From the necessary equality of the 
resulting forces acting in any section, 
C = r,C' = T'andC -C' = T-r. 
The algebraic sum of the forces 
acting below the neutral surface is a 
force T — T\ applied at the center 
of gravity of the steel. This force 
tends to shear the concrete on lany 
horizontal section between the steel and the neutral sur- 
face. The stress resisting this force is distributed over 
an area equal to the width multipHed by the length of the 




Fig. 9L 



REINFORCED CONCRETE 189 

section, hence the unit shearing stress parallel to the 
neutral surface between the steel and the neutral surface 
will be 

>s. =^^— ^. (1) 

oal 

For equihbrium the moment of the couple {C — C'){T — T') 
must be equal to Vdl, where V is the vertical shear for the 
section considered. The arm of the couple is jd, hence 

Vdl = (T - r)jd (2) , 

Ehminating T — T' between the equations (1) and (2) 

Sn=^.- (3) 

hjd 

Equation (3) holds true for T-section beams as well as 
for rectangular ones when h is the width of the section on 
the horizontal plane for which Sn is the unit shearing stress. 

Above the neutral surface the horizontal unit shearing 
stress is always less than below that surface, hence the 
maximum horizontal unit shearing stress occurs below the 
neutral axis in a section where F is a maximum, and will 
have the greatest possible value for that stress when h is 
the least width of that section. 

Art. 92. Bond Stress. 

Since the theory for reinforced beams assumes that there 
is a perfect bond between the steel and the concrete, the 
difference between the total tension in the steel at any 
two sections of a beam must be equal to the resistance that 
the concrete surrounding the steel between those sections 
offers to prevent any relative motion between the steel 
and the concrete. 

The value of the bond stress per square inch of horizontal 
rod surface may be found as follows: Let T and T^ be the 



190 MECHANICS OF MATERIALS 

total tensions in the steel at two sections dl apart, then 

T — T' 

is the shearing or bond strength per inch of length 

dl 

of the beam. From the equality of moments (Fig. 91), 

Vdl = (T - T')jd. 

T — T' V 

It follows that — = — , which gives the value of 

di jd 

the bond stress per unit of length of the beam at a section 

of the beam where the vertical shear is V. This stress, 

divided by the sum of the perimeters of all the bars, is the 

bond stress per square inch of rod surface. 

If u is the bond stress/sq. in. and 2o is the sum of 

the perimeters, then 

V 

u = — . 

Zojd 

Art. 93. Diagonal Stress. 

In Chapter VII, Article 80, it was proven that in any 
section of a beam there was a maximum tensile stress 
whose value was determined by 




2 S 
acting at an angle 6 such that tan 2 6 = —~. 

In reinforced concrete beams the horizontal tensile stress 
below the neutral axis is assumed to be zero, and since 
Ss = Sh, it follows that Sn, equal in magnitude to Sn, is a 
maximum unit tensile stress which acts at an angle of 
45 degrees with the axis of the beam. 

Since the value of this diagonal tensile stress depends 
on the magnitude of the horizontal shearing stress, the 
greatest value of Sn will be found in a section at or near 
to the supports where the vertical shear is a maximum. 
When the failure of the beam is due to the diagonal tension, 



REINFORCED CONCRETE 191 

the line of fracture generally starts on the plane of the 
reinforcement at or near to the supports and extends 
towards the center of the beam, making an angle of approx- 
imately 45 degrees with the axis of the beam. 

The value for Sn = Sn was obtained on the assumption 
that the horizontal tensile stress was zero. As there is 
some tensile stress in the concrete at all sections of a beam, 
it must always be remembered that the true value of Sn 
is always greater than Sn- 

While in some beams the value of Sn never becomes 
large enough to be a dangerous stress in the concrete, a 
great many of the failures of reinforced concrete beams are 
due to the neglect to provide proper reinforcement against 
the diagonal stress. 

To determine the area of the steel required to aid the 
concrete in resisting the diagonal tension, it will be neces- 
sary to make some assumptions as to the distribution of 
the stress between the steel and the concrete and the 
probable value of the vertical and horizontal components 
of the diagonal stress. 

Regarding the distribution of the stress one very common 
assumption considers that the steel carries f and the con- 
crete ^ of the diagonal stress, no steel being required until 
that stress exceeds 35 to 40 Ib./sq. in. Another considers 
that the concrete will carry 35 to 40 Ib./sq. in. and that 
the balance is to be carried in the steel. 

If the unit tensile stress in the concrete is hmited to 35 
to 40 Ib./sq. in. and the maximum value of Sn for any 
beam, regardless of the amount of steel reinforcement, lies 
between 105 and 120 Ib./sq. in., each of the assumptions 
will give the same result when Sn has the maximum value, 
but for all values of Sn less than the maximum the former 
will require more steel than the latter. The present ten- 



192 MECHANICS OF MATERIALS 

dency seems to be towards the use of the one that requires 
the most steel. 

Similarly while the common practice assumes that ver- 
tical and horizontal components of the diagonal stress are 
each equal in magnitude to S^, one authority gives the 

V V 

value of Sn as -— and another as -— . Since the average 
jbd od 

value of j is |, these two assumptions will give the same 

result if the maximum stresses in the latter case are f of 

those used in the former case. 

Consider a short length of a beam, s, lying between 

two sections where the vertical shears are Vi and F2. If 

Vi + V2 
the section is short then Va = may be taken as 

value of a uniform vertical shear extending over the small 
bngth s. Taking the vertical component of the diagonal 

Va VaS 

stress as numerically equal to Sn = -rr-.j —rr is the total 

Jbd jd 

vertical stress exerted over "a section sXh in area. Of 

this total vertical force the steel is to carry |, therefore 

the force that the steel has to carry is given by ~. 

3 jd 

When the steel reinforcement is placed with the axis of 
the bars perpendicular to the neutral surface of the beam, 
then the area of steel required in the length s is 

2_VaS 

3 jdfs 

When the axis of the bars is perpendicular to the plane 
of the diagonal fracture the effective area of the steel to 
resist vertical force becomes 

^ 2^ Va cos 45°s 
"^ " 3 jdfs ' 



REINFORCED CONCRETE 193 

The minimum number of bars that should be used in the 
distance s is given by the experimentally determined fact 
that the bars are not effective if the distance between any 
two bars is greater than f d. 

The number of bars in excess of this minimum is deter- 
mined by the bond strength of the bars, and other con- 
ditions depending on practice. 

Art. 94. Columns. 

A concrete column may be reinforced by steel bars 
placed parallel to the axis of the column, or by steel bands 
or wire being wound spirally around the column. 

In the former case both the steel and the concrete 
sustain the same deformation and the laws for columns of 
two materials hold true. Failure of the column would 
probably take place when the elastic limit for the concrete 
is reached. 

When the reinforcing steel is wound spirally around the 
column, the steel cannot carry any of the compressive load 
directly. When a bar sustains an axial compressive force 
there is an increase in the lateral dimensions. If this 
deformation, could be prevented the allowable compressive 
load would be greatly increased. Since for compressive 
stresses within the elastic limit of the concrete there is 
only a slight lateral deformation, it is evident that the 
elastic strength of the column is but slightly increased by 
the use of spiral reinforcement. The steel bands will 
undoubtedly aid in preventing failure of the column when 
the elastic strength of the column is exceeded, but the 
longitudinal deformation will be excessive. To be effect- 
ive the clear spacing between the spiral bands must not 
exceed one-quarter of the diameter of the enclosed column. 

The effective area of a reinforced concrete column is 



194 



MECHANICS OF MATERIALS 



the area enclosed by the reinforcing steel. Any concrete 
there may be that is outside of the steel is placed there for 
the protection of the steel. Since the use of spiral banding 
may increase the ultimate load that a column can support, 



r» \ 

— ^ 

===== =^ 



1ft' a", .___ ^ 


O 


n n 










t 








» 








1 









S-^a'^Rods tied with No. 16 Wire to Mild Steel Hoops spaced TO'cto c 





Spacing Bars 
jTHigh Carbon Steel Spiral with I'Pitch 



Spacing Bars 




_ J^'High Cart»h Steel Spiral vyith TPitch. 
^^''♦Rods Beo to Spiral with No. 16 Wire at ro** Intervals 

Fig. 94. 



Spadng Bars 



and has but little effect on the elastic strength of the column, 
the allowable unit stress in the concrete in a column having 
both spiral and longitudinal reinforcement is higher than 
when the longitudinal bars are used alone. The common 
practice seems to tend towards the use of longitudinal 
reinforcing bars held in position by ample spiral banding, 



REINFORCED CONCRETE 195 

and to take no account of the spiral reinforcement in so far 
as it is effective in aiding the concrete to carry the safe 
load. 

Art. 95. Deflection of Reinforced Concrete Beams. 

Its complex character and the uncertainty as to the dis- 
tribution of the stresses in the concrete, render the deriva- 
tion of a theoretically correct formula for the deflection of 
a reinforced concrete beam practically impossible. It is, 
however, possible to obtain an expressiou for the deflection 
which while theoretical in form contains an empirically 
determined constant. When the conditions under which 
the beam is to be used are approximately the same as those 
under which the constant was determined, the expression 
will give very satisfactory results. 

The following discussion is due to Turneaure and 
Maurer,* with slight change in notation. 

The form of the expression is the same as that for beams 
of one material throughout. This for a rectangular con- 
crete beam is 



^EJ 



(1) 



To find the value of I for a reinforced beam, the area 
of the steel will be considered as increased n times without 
changing the position of its center of gravity relative to the 
neutral axis of the section and the steel replaced by this 
area of concrete. This gives an all concrete section with 
the same theoretical strength as the given reinforced 
section. The tensile stress in the concrete below the 
neutral axis affects the amount of deflection and should 
not be neglected. In the absence of any information as 
to the depth below the axis that the tensile stress is effect- 
* Principles of Reinforced Concrete, Turneaure and Maurer. 



196 MECHANICS OF MATERIALS 

ive it is assumed that the entire area between the steel and 
neutral axis carries a tensile stress. 

This assumption changes the value of k from that given 
in Art. 89 to 

, ^ ^ + 2np 
"~ 2 + 2 np* 

Referring to Fig. 89 it will be seen that the moments of 
iiiertia of the areas in tension and compression relative to 

the neutral axis are/— (/cd)^ for concrete area in compression, 

6 r p 

— (1 — k)d\ for the concrete area in tension, and for 

the changed steel area nphd[(l — k)d]^ approximately, 
hence the moment of inertia of the entire section is, 

/ = i-| A;3 + (1 _ /c)3 4_ 3 pri(l - k)AhdK 

If the coefficient of hd^ is represented by cf), substituting 

Es 
hd^ <^ for 7, and — for Ec in equation (1), the expression 

n 

for the deflection of a rectangular reinforced concrete 
beam becomes, 

^ = ^^rL (2) 

The value of n * to be used in the above formula, while 
theoretically the ratio between the moduli of elasticity 
for steel and concrete, should be chosen so that for the same 
beams, the calculated value of A will agree with the experi- 
mentally determined value of the deflection. 

The variation in the distribution of the tensile stress in 

* Turneaure and Maurer recommend that 8 or 10 be used for n in 
the formulas for deflection. 



REINFORCED CONCRETE 197 

the concrete for various intensities of load on any beam 
renders it unlikely that any single value of n could satisfy 
all variations of load, hence the formula as derived should 
be used to determine the deflection under the maximum 
safe load for the beam only. 

Following the same line of reasoning and considering 
the areas of the web above and below the neutral axis as 
effective in carrying compression and tension respectively 
the maximum deflection for a reinforced concrete T-beam 
becomes, 

in which the value of d is given by, 



and of k by, 

k = 


, 1 \b' v/tv ,(t \n 




,6' V t\ t 



Art. 96. Use of Formula. 

While the assumptions made when the relation between 
the external forces and the induced unit stresses in rein- 
forced concrete bars were derived are approximately true, 
none of them can be considered as exactly so. Hence the 
formulas derived for the various unit stresses in reinforced 
concrete bars should be considered as theoretical in form 
only. Each expression contains one or more constants of 
materials such as the safe unit stress modulus of elasticity, 
etc. 



198 MECHANICS OF MATERIALS 

When the value of at least one of these constants in 
each expression is determined so that the result as cal- 
culated by the formula will agree with those obtained by 
experiments on similar bars, the expressions become empiri- 
cal and their use should be limited to the range of experi- 
mental work. Considering the complex character of con- 
crete, and the many conditions that may affect the strength 
of reinforced concrete over which the designer has no direct 
control, it is to be expected that there may be considerable 
difference in the designs offered for a reinforced concrete 
structure by reputable engineers, but such discrepancies 
will be greatly reduced as our knowledge of the action of 
reinforced structures under load is increased. 

In 1909 a joint committee from the American Society 
of Civil Engineers, The American Society for Testing 
Materials, the American Railway Engineering and Main- 
tenance of Way Association, and the Association of Ameri- 
can Portland Cement Manufacturers, after an exhaustive 
study of the then existing experiments on reinforced con- 
crete, made certain recommendations regarding the proper 
values for material constants to be used, and when these 
values are used in connection with the formulas as derived 
in this chapter the design will represent the average prac- 
tice of that day.* 

PROBLEMS 

Es 
■ The values of — and the allowable unit stresses to he 

Ec 
used in the solution of the following problems will be found in 
the appendix under " The Recommendations of the Joint Com- 
mittee,^' unless otherwise stated. 

* An abstract is given in the appendix for the use of the student 
in solving the problems given in the text. 



REINFORCED CONCRETE 199 

1. A reinforced concrete slab 6 in. deep, 4 ft. wide and 
6 ft. long, has 2 sq. in. of steel reinforcement placed 2 in. 
from the lower surface of the beam. Find the maximum 
horizontal unit tensile and compressive stresses in the steel 
and concrete and the greatest uniform load that can be 
carried. Consider the weight of the slab as part of the 
uniform load. 

2. Find the ideal sectional area of the horizontal steel 
reinforcement for a slab of the same dimensions as the one 
given in problem 1. Determine the uniform load that 
may be carried with the ideal amount of reinforcement. 

3. Find the area of the steel reinforcement needed in 
a concrete beam 10 ft. long, 12 in. wide and 12 in. deep to 
the center of the steel reinforcement to carry a load of 
6900 lb. at the middle of the span. Consider horizontal 
tensile and compressive stresses only. 

4. Find the maximum horizontal unit shearing stress in 
the beam as given in problem 3. Does this beam need 
any reinforcement against diagonal tension? 

5. A concrete beam 8 in. wide and 18 in. deep with 3 
f -in. square bars placed 2 in. from the lower surface of the 
beam, is used to cover a span of 12 ft. and support a 
uniform load of 13,250 lb. Find the maximum bond 
stress and the greatest horizontal unit stresses in tension 
and compression and shear. 

6. Find the sectional area of the vertical reinforcement 
needed in the beam as given in problem 5 in a distance 
of 16 in. from the end of the beam. Assume the allowable 
unit stress in the vertical steel as 12,000 Ib./sq. in. 

7. The unit tensile stress in a steel rod J in. diameter is 
16,000 Ib./sq. in. Find length that this rod should be 



200 MECHANICS OF MATERIALS 

embedded in concrete if the bond stress is limited to 
100 Ib./sq. in. of rod surface. 

8. Find the maximum horizontal unit shearing stresses 
in the beams given in problems 1 and 2. 

Assume that the length of the columns in the following 
problems is less than 15 diameters. 

9. A concrete column 16 in. effective diameter is rein- 
forced by eight f -in. steel rods parallel to the axis and held 
in place by being connected to a No. 16 wire wound spirally 
around the longitudinal bars, spaced 4 in. from center to 
center. Find the safe load for the column. 

10. Determine the load that could be safely carried on 
the column given in problem 9, assuming that the longi- 
tudinal reinforcement was the same and that there were but 
four bands used to hold the steel in position. 

11. Find the sectional area of the longitudinal reinforce- 
ment needed in order that a concrete column 12 in. effective 
diameter may safely carry a load of 102,000 lb. Consider 
that there is ample spiral reinforcement and length = 12 ft. 

12. Find the probable amount of shortening of the 
column given in problem 11 when the given load is 
applied. 

13. Find the maximum deflection for the beams given 
in problems 3 and 5. 

14. Given a T-section concrete beam, flange 12 in. thick 
by 48 in. wide, web 18 in. wide, reinforced by five if-in. 
and seven IJ-in. diameter steel rods placed horizontally 
in three rows with their center of gravity 47 in. from the 
top of the flange. 

Find: 

(a) The maximum horizontal unit stresses in the steel 
and concrete in tension, compression and shear. 



REINFORCED CONCRETE 201 

(6) The bond stress per unit of rod surface. 

(c) Consider a length of the beam extending from the 
edge of the support towards the center of the span for a 
distance of 2 ft. Find the sectional area of the vertical 
steel rods required to reinforce this part of the beam 
against diagonal tension, allowing a unit stress in the steel 
of 12,000 Ib./sq. in. 

(d) Find the deflection of the beam under the given load. 



EXPLANATION OF TABLES 

Table I. Fundamental Formulas. 

Table II. Derived Formulas. 

The numbers following each expression refer to the chapter and 
article in which the formula was derived. 

Table III. Prop^erties of Beams. 

The columns 1 and 2 give the relative strengths and stiffness of the 
various kinds of beams of the same length and shape. Columns 3 to 6 
are the expressions for Maximum Vertical Shear, Bending Moment, 
Unit Stress, and deflection of the various beams under uniform and 
single concentrated loads. Columns 7 and 8 give the values of a. and /3 
for the various beams; for a description of these symbols see: for a, 
Art. 48; i8. Art. 64. 

Table IV. Constants of Materials. 

This table has been compiled solely for the use of the student in 
solving the problems in the text. As all the constants are liable to 
considerable variation, it should not be used in the design of a structure 
that is to be built. 

Table V. Properties of Sections. 

In the rectangular sections d is the dimension in the direction of 
bending. In the hollow sections dx and 6i are the inside dimensions. 

Tables VI and VII. Properties of I and Channel Beams. — 

Cambria Steel. 

These tables have been inserted for the convenience of the student. 
As every engineer should own some of the trade books giving the 
properties of the various steel shapes, he will prefer to get his data 
first hand. 

Table VIII. A partial list of the recommendations of the Joint 
Committee regarding the allowable unit stresses in reinforced concrete 
beams and columns. 

202 



FUNDAMENTAL .J^ORMULAS 203 



FUNDAMENTAL FORMULAS.— TABLE I 

Tension, Compression, and Shear 

(a) P = AS. Chap. I, Art. 4. 
Applies to all cases of uniformly distributed stress. 

Modulus of Elasticity for Tension and Compression 

(b) E = - = ^. Chap. I, Art. 11. 

e Ae 

Applies to all problems where the unit stress in tension or com- 
pression is within the elastic limit. 

Beams. — Vertical Shear 

(c) V=AS. Chap. Ill, Art. 39. 
True for all values of S. 

Bending Moment 

(d) M = ^. Chap. Ill, Art. 40. 

c 

Applies to all problems where the value of S is within the elastic 
limit. 



(/) 

Applies to all problems where the value of S is within the elastic 
limit. 

Equation of Elastic Curve 

EI^ = M. Chap. V, Art. 63. 



Twisting Moment in 


Shafts 


> 


p _SJ 
c 




Chap. IV, Art. 53. 


p_Sl_Ppl 
Oc 6J' 




Chap. IV, Art. 56. 



204 



MECHANICS OF MFATERIALS 



DERIVED FORMULAS. — TABLE II 
Strength of Bars of Uniform Strength 

logio ^0- Chap. II, Art. 23. 



w 



logioA = 0.434- y + 



Thickness op Steam and WaterJ Pipes, Cylinders, etc 

Thin pipes : 

Longitudinal ruptures. 

RD = 2 St. 

Circumferential ruptures. 

RD = 4: St. 
Thick pipes : 

Longitudinal ruptures. 

RDj^ = 2 St. 



Chap. II, Art. 24. 
Chap. II, Art. 24. 



Strength of Riveted Joints 
Tension in plate. t{p — d)St = Pf 

cird^ 



Shear on rivet. 



4 



5, = P., 



Chap. II, Art. 25. 

Chap. II, Art. 30. 
Chap. II, Art. 30. 

Chap. II, Art. 30. 
Chap. IV, Art. 58. 



Compression on rivet or plate. c^tdSc = P^ 

Horse Power of Shafts 

„ PpN , . SsJN 

H= —^ — (approx.) = \—^ ', 

63,000^^^ ^ 53,000 c 

Shaft CouplijJ^gs 
Diameter of bolts. Pp = n~- SJ'h (ai)prox.). Chap. IV, Art. 59. 



Helical Springs 



Strength. 
Deflection, 



S = 



P = l^Ss. 
SD 

8PZ)3 s.ttD^ 



Fd^ 



Fd 



Chap. IV, Art. 61. 



DERIVED FORMULAS.— TABLE II 205 



Continuous Beams. — Three-moment Equation 
N^li + 2N2 (li + I2) + N^l^ = - !f!A!_±J£2^. Chap. V, Art. 67. 



Long Columns 



Round ends. 



Euler. P = ^^. Chap. VI, Art. 70. 

Rankine. P = ^^' ., . Chap. VI, Art. 73. 



Square ends. 

Euler. p = 4:^I^. Chap. VI, Art. 71. 

Rankine. P = -^^^' Chap. VI, Art. 73. 

1 + <^1 

Round and square ends. 

Euler. P = ^ ^^. Chap. VI, Art. 72. 

Rankine. P = ^^^"7^ • Chap. VI, Art. 73. 

9 ^2 

Combined Stresses 
Tension or compression with ben'ding. 



^^ ^^a P/2' 



a 1 1 -^ 

^ E 

P Mc 
or 5n, = - + -— (approx.). 

^ 1 

Tension or compression combined with shear. 

Max. shear. Sp =^ ± V,S7TT^ Chap. VII, Art. 80. 

Max. tension or , 

compression . «» = ^ « ± -^SjTlS^. Chap. VII, Art. 80. 

Horizontal Shearing Stress in Beams 

V C V 

'Sfft = — I yd A = —aici. Chap. VII, Art. 81. 
lb J lb 



206 MECHANICS OF MATERIALS 

Compound Coujmns 
Relative unit stress induced. 



^=-^...=-^. Chap. VIII, Art. 85. 

Mix JOJ2 J^n 



Relation between the partial loads. 

Pih Pih. Pnln 

AiEi A2E2 AnEn 

Totalload = Pi + P2 . . . + Pn- 

Compound Beams 
Relative unit stress induced. 

« fik"^ a. fih^ a fnl\ 



Chap. VIII, Art. 85. 



|8 EiCi ^ E2C2 fi EnCn 

Relation between the partial loads. 

Wik^ ^ W2II _ Wnl'n 

Total load = W1+W2 . . . Wn. 



. Chap. VIII, Art. 86. 



Chap. VIII, Art. 86. 



Reinforced Concrete 



Rectangular Beams. 



Jc = = pn 

1+4 

nfc 



Jl+--l-l|. 
_M pn J 



1 k 

Sc\ nfc/ 
M = fs pbdjd = —khdjd.] 

Wl^ n 

A = • -. 

pEsbd^ <t> 



DERIVED FORMULAS.— TABLE II 207 



<!> = -[k' + (1 - /c)3 + 3 pn(l - A;)2]. Chap. IX, Art. 95. 

o 



For deflection only. 

1 +2pn 



k = 



2 +2pn 



T-beams. 

vn -f 
1 ^ 2\d, 
k = — = . Chap. IX, Art. 90. 

1 + — - pn + — 

nfc a 

i 
3/c - 2 - 

d 

z = -, jd = d - z. Chap. IX, Art. 90. 

2/c - - 
d 

M =fs pbdjd =f4^ - ^j^l ^h'd- Chap. IX, Art. 90. 

Wl^ n 
A =^^^3-. Chap. XI, Art. 95. 

Chap. IX, Art. 95. 
For deflection only. 

b b\d) '^ \d) \ 

— \r , • Chap. IX, Art. 95. 

6 bit 

^ b bd d 
Horizontal Shearing Stress (all beams). 

V 
Sh = — . Chap. IX, Art. 91. 



= r[ 



k ^ 



jbd 
Bond Stress (horizontal bars in beams) 



Columns. 

P = Acfc + nAsfc 



V 
u = -— . Chap. IX, Art. 92. 



P 

/c = -; — TT"- C^ap. IX, Art. 94. 

Ac + Agn 



208 



MECHANICS OF MATERIALS 






02 

W 

M 
O 

H 

o 

Ph 



00 00. 


CO 


00 


00 


00 

CO 


lO 


CO 
00 
r-i 


(M 


00 

CO 


t- s 


tH 


(M ■ 


^ 


00 


00 


00 


(M 


6 

Max. 
Deflec- 
tion 


5 


CO 


eo 


GO 




00 


eo 


00 
CO 


CO 

5 


CD 
00 

T— ( 


eo 


CM 
Oi 

T— 1 


eo 


00 
CO 


5 

Max. 

Tensile or 
Compressive 
Unit Stress 




^ 




CM 


^ 

^ 




^ 

k 


00 




00. 




00 




rH 


4 

Max. 
Bending 
Moment 


s 


^ CM 




s °° 




^ CO 


Ss 


3 

Max. 

Vertical 
Shear 


fe 


fe 


^ CM 


^|oi 




00 


^^ 


^ CM 


2 

Eelative 

Stiffness 


1— 1 


ooico 


CD 


00 
(M O 


(M 

CO 


CD 


00 
CM 

l-H 


1 

Relative 
Strength ■ 


iH 


(M 


rtH 


00 


00 


00 


CM 

T— ( 


Q 
<i 
O 

fa 
o 

M 


h 
H 

r 

r 
r 

r 

F 


rH 

d 

O 

— 1 

-H 
rH 


h 
t- 

r 
r 
r 


1-1 

s 

O 

H 

rH 

c3 
O 

—I 

^1 

> 

— t 

pi 

Z) 


<X> 

l-H 
'd 

'd 

• rH 

B 
'd 

CD 

o 

1 — 1 

C3 
<U 

rO 

<D 

r-H 

a 

• rH 


4^ 
■^ 

'a 

'r-l 

o 

%-H 

'3 

O 

l-H 

a 

O) 
'Ph 

a 


C3 

'd 
-p 
o 

Ph 
Ph 

o 

H-=> 
C« 

M 

a 

:« 
<V 

m 


o 

i-H 

a 
1 

PI 
o 


1i 

<D 

C3 

i=l 
<V 

O 
-4-= 

CD 

g 


r^ 
'r-l 

a 


• fi 

*a 

O 

1— ( 

rCj 

o 

H-3 

M 

ca 
S 

c3 

p:; 


rb 

a 



AVERAGE CONSTANTS OF MATERIALS. — TABLE V 209 



0) 




CO 
00 




O 

d 
I— 1 


d 

1— 1 


d 

1—1 


O 
Ol 


CO 
CO 


T-H 

CO 




o 


o 


o 

00 


o 


o 


CO 
05 


o 

CO 

T-l 


o 

T-H 


Shearing 

Modulus of 

Elasticity 

E 

Lbs./Sq. In. 


o 
o 
o 

o 
1—1 


o 

o 
o 
o 
o 
o 


O 

o 

o 

<s 

o 
o 

o 

T— 1 


o 
o 
o 

o 

o 
o 

of 

l—t 


o 
o 
o 

d~ 
o 
o 

1—1 














Modulus of 

Elasticity 

E 

Lbs./Sq. In. 


o 
o 
o 

o 

1— r 


o 
o 
o 
cT 

§ 

r-( 


o 
o 
o 

d 
o 
o 

oo" 

(M 


o 
o 
o 

d' 
o 
o 

<s 

CO 


o 
o 
o 

o 
o 
o 

<s 

CO 










o 
o 
o 

o 


w 

EH 

'A 

M 
H 
02 

W 

<J . 

s 

H 
I-] 


1— 1 
yA 


a 

'S o 

o ^ 

a '^ 
o 


O 

o 
o 


o 
o 

o 

•<* 


o 
o 

o 


o 
o 
o 


o 
o 
o 

1—1 


o 

o 






Sod 


O 
O 

o 

CO 


o 
o 
o 
o 

00 


o 
o 
o 


o 
o 
o 
o 

CO 


o 
o 
o 

<s 

o 


o 
o 
o 

CO 


o 
o 
o 

CD 


o 


a 

1-1 


o 

■ o 

o 

00 


o 
o 
o 

CM 


o 
o 


o 
o 
o 

d~ 

CD 


o 
o 
o 

d~ 
o 

T— 1 


o 
o 






(N 


Elastic 

Limit 

EL 

Lbs./Sq. In. 


o 
o 
o 

CO 


o 
o 
o 

CO 


o ' 

o 

o 


o 
o 
o 

d^ 

CO 


o 
o 
o 

o 

00 














3 


u 

s 

H 


o 

m 


o 

u 

ho 

o 

u 


1 — 1 
Oi 
<D 

u 
o 


r— I 
CD 
O) 

CO 

w 


PC 




a 

s 




a 
a 

C 

c 





210 



MECHANICS OF MATERIALS 






"A 

o 

H 
O 

O 
H 

o 

?^ 









(N 




















'^ 


'^ 








•* 








CO 














00 

►J 

to 

O 


^.1 o 


+ 

eo 


+ 

> 

CD 








03 rH 


I 


rH 


iz; 


























o 
















Si o 




S^Ico 




^ 


o 


■* 1 

1 


O 














•«»< 


r-( 














'Q 














^^ 




1 


^ t^ 02 








^0 










55g 

<< y s 


C4 


SL|C^ 


(M IfM 


1 


1 


^ 1 CO 




CD 


«* 


^It-H 


'^It-I 


1 




-^I^ 


+ 


rH 










^ 


















rO 








CO 






























p 

^ 




















>< 

H 


S 


+ 

CO 


CM 


^1^ 






1 CO 


1 


(M 

CO 


C5 




'^1 












<) 














^ 














« 














w 














S5 

t-H 








CO 






■* 




1^ 








^ 






■^ 




o 


K| 


^ (M 


^I'^l 


fO 


3^ 


^ -^ 


1 


'^ 


iz; 




I-o ""I 


^ |tH 


1 


i—l 


^ CO 


■^ 


CO 


H 

s 

^ 








CO 






^ 




s 








rO 






V 












i-» 




IM 












-«. 




'^^ 




<1 








fC) 


<M 








^ 






1 


1^ 


1 


rtH 






^3 










(Si 




X. 










o 




4^ 




03 






s 




^ ^ 








s 




o) a^ 




bO 






H 




-r^ "^ 




rt 






OS 




^ \. 




ce 

-t-3 
O 




0) 


' § 




^1 


I 


1— 1 


1—1 

• l-H 


H 




O +5 


1:3 


u 


o 


OQ 








^ 




^ 




^ 2^ 


T3 


o^ 


nd 


o^ 






•'-' r\ 


• iH 






1— 1 






'7? '^ 


r-^ 


o 


■-H 


o 






O 
CO 




o 
CZ2 


tl 


\ 1 


o 


^ 


J 11 



PROPERTIES OF STANDARD I-BEAMS 



211 



PROPERTIES OF STANDARD I-BEAMS.— TABLE VI. 

Axis 1 — 1 Passes through Center of Gravity Perpendicular to Web. 
Axis 2 — 2 Passes through Center of Gravity Parallel to Web. 



1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


11 










xi 




s 




a 


2 


a 


(J 

Si 

g 


0^ 


o 
o 






6 


"■+3 

a 




o 
a 

>5 


a 


o 
'■+3 

c3 
(-, 
>» 


3 


o 


- tj 
ft 




o 

m 
a: 
<0 




C|_l • 


a 








a 
.2 

u 


ft 






a 




1^ », 


II 


2 n 

1< 



a; 03 


3 2 


0) 
. CO 


Q 


^ 


< 


H 


S 


s 


4) 
02 


« 


^ 


05 




d 


A 


t 


b 


I 


s 


r 


I' 


r' 




Ins. 


Lbs. 


Sq. in. 


Ins. 


Ins. 


Ins.4 


Ins.3 


Ins. 


Ins.4 


Ins. 


B5 


3 


5.5 


1.63 


.17 


2.33 


2.5 


1.7 


1.23 


.46 


.53 


** 


" 


6.5 


1.91 


.26 


2.42 


2.7 


1.8 


1.19 


.53 


.52 


** 


" 


7.5 


2.21 


.36 


2.52 


2.9 


1.9 


1.15 


.60 


.52 


B9 


4 


7.5 


2.21 


.19 


2.66 


6.0 


3.0 


1.64 


.77 


.59 


" 


" 


8.5 


2.50 


.26 


2.73 


6.4 


3.2 


1.59 


.85 


.58 


" 


" 


9.5 


2.79 


.34 


2.81 


6.7 


3.4 


1.54 


.93 


.58 


* 


" 


10.5 


3.09 


.41 


2.88 


7.1 


3.6 


1.52 


1.01 


.57 


B13 


5 


9.75 


2.87 


.21 


3.00 


12.1 


4.8 


2.05 


1.23 


.65 


** 


" 


12.25 


3.60 


.36 


3.15 


13.6 


5.4 


1.94 


1.45 


.63 


** 


** 


14.75 


4.34 


.50 


3.29 


15.1 


6.1 


1.87 


1.70 


.63 


B17 


6 


12.25 


3.61 


.23 


3.33 


21.8 


7.3 


2.46 


1.85 


.72 


" 


" 


14.75 


4.34 


.35 


3.45 


24.0 


8.0 


2.35 


2.09 


.69 


" 


" 


17.25 


5.07 


.47 


3.57 


26.2 


8.7 


2.27 


2.36 


.68 


B21 


7 


15.0 


4.42 


.25 


3.66 


36.2 


10.4 


2.86 


2.67 


.78 


" 


*' 


17.5 


5.15 


.35 


3.76 


39.2 


11.2 


2.76 


2.94 


.76 


** 


** 


20.0 


5.88 


.46 


3.87 


-42.2 


12.1 


2.68 


3.24 


.74 


B25 


8 


17.75 


5.33 ■■ 


.27 


4.00 


56.9 


14.2 


3.27 


3.78 


.84 


" 


" 


20.. 25 


5.96 


.35 


4.08 


60.2 


15.0 


3.18 


4.04 


.82 


" 


" 


22.75 


6.69 


.44 


4.17 


64.1 


16.0 


3.10 


4.36 


.81 


** 


** 


25.25 


7.43 


.53 


4.26 


68.0 


17.0 


3.03 


4.71 


.80 


B29 


9 


21.0 


6.31 


.29 


4.33 


84.9 


18.9 


3.67 


5.16 


.90 


" 


" 


25.0 


7.35 


.41 


4.45 


91.9 


20.4 


3.54 


5.65 


.88 


" 


" 


30.0 


8.82 


.57 


4.61 


101.9 


22.6 


3.40 


6.42 


.85 


** 


** 


35.0 


10.29 


.73 


4.77 


111.8 


24.8 


3.30 


7.31 


.84 


B33 


10 


25.0 


7.37 


.31 


4.66 


122.1 


24.4 


4.07 


6.89 


.97 


*' 


" 


30.0 


8.82 


.45 


4.80 


134.2 


26.8 


3.90 


7.65 


.93 


** 


" 


35.0 


10.29 


.60 


4.95 


146.4 


29.3 


3.77 


8.52 


.91 


" 




40.0 


11.76 


.75 


5.10 


158.7 


31.7 


3.67 


9.50 


.90 




12 


31.5 


9.26 


.35 


5.00 


215.8 


36.0 


4.83 


9.50 


1.01 


B 41 


' ' 


35.0 


10.29 


.44 


5,09 


228.3 


38.0 


4.71 


10.07 


.99 


" 




40.0 


11.76 


.56 


4.21 


245.9 


41.0 


4.57 


10.95 


.96 


B 53 


15 


42.0 


12.48 


.41 


5.50 


441.8 


58.9 


5.95 


14.62 


1.08 


** 


" 


45.0 


13 . 24 


.46 


5.55 


455.8 


60.8 


5.87 


15.09 


1.07 


** 


" 


50.0 


14.71 


.56 


5.65 


483.4 


64.5 


5.73 


16.04 


1.04 


** 


** 


55.0 


16.18 


.66 


5.75 


511.0 


68.1 


5.62 


17.06 


1.03 






60.0 


17.65 


.75 


5.84 


538.6 


71.8 


5.52 


18.17 


1.01 



212 



MECHANICS OF MATERIALS 



PROPERTIES OF STANDARD CHANNELS.— TABLE VII. 

Axis 1-1 Passes through Center of Gravity Perpendicular to Web. 
Axis 2-2 Passes through Center of Gravity Parallel to Web. 



1 


2 


3 


4 


5 


6 


7 


8 


9 


10 


11 


12 


13 


a 


'S 
a 
a 

03 

O 
o 


4J 

o 
o 

ft 


d 
o 
'•+3 

a) 
CO 


m 


a> 
bC 

c3 

S 

o 


'■+3 

a 




.2 

'■+3 
u 


I— 1 






o 

03 
u 
>> . 


4; Bji 

£3 >-> 


a 
o 


ja 




o 




j:i 


ID 03 


§■2 


S.2 


0) 2 


§.2 


3.2 


sO'2 




-fj 


bc 


c^ 


Q 


-tj 




• rt X 


^ ^ 




■t X 


•S X 


K ^ 




a 


'53 


0) 


'jg 


TJ 


t< 


03^ 


t< 


'§-< 


2oO 


to 


P 


^ 


< 


H 


g 


s 




rt 


^ 


m 


« 


Q 




d 


A 


t 


b 


I 


S 


r 


I' 


S' 


r 


X 




Ins. 


Lbs. 


Sq.Ins. 


Ins. 


Ins. 


Ins.4 


Ins.3 


Ins. 


Ins.* 


Ins.3 


Ins. 


Ins. 


C5 


3 


4.00 


1.19 


.17 


1.41 


1.6 


1.1 


1.17 


.20 


.21 


.41 


.44 


" 


" 


5.00 


1.47 


.26 


1.50 


1.8 


1.2 


1.12 


.25 


.24 


.41 


.44 


" 


4t 


6.00 


1.76 


.36 


1.60 


2.1 


1.4 


1.08 


.31 


.27 


.42 


.46 


C9 


4 


5.25 


1.55 


.18 


1.58 


3.8 


1.9 


1.56 


.32 


.29 


.45 


.46 


" 


" 


6.25 


1.84 


.25 


1.65 


4.2 


2.1 


1.51 


.38 


.32 


.45 


.46 


" 


" 


7.25 


2.13 


.33 


1.73 


4.6 


2.3 


1.46 


.44 


.35 


.46 


.46 


C13 


5 


6.50 


1.95 


.19 


1.75 


7.4 


3.0 


1.95 


.48 


.38 


.50 


.49 


" 


" 


9.00 


2.65 


.33 


1.89 


8.9 


3.5 


1.83 


.64 


.45 


.49 


.48 


(■ 


" 


11.50 


3.38 


.48 


2.04 


10.4 


4.2 


1.75 


.82 


.54 


.49 


.51 


C17 


6 


8.00 


2.38 


.20 


1.92 


13.0 


4.3 


2.34 


.70 


.50 


.54 


.52 


<■ 


" 


10.50 


3.09 


.32 


2.04 


15.1 


5.0 


2.21 


.88 


.57 


.53 


.50 


•• 


" 


13.00 


3.82 


.44 


2.16 


17.3 


5.8 


2.13 


1.07 


.65 


.53 


.52 


<i 


" 


15.50 


4.56 


.56 


2.28 


19.5 


6.5 


2.07 


1.28 


.74 


.53 


.55 


C21 


7 


9.75 


2.85 


.21 


2.09 


21.1 


6.0 


2.72 


.98 


.63 


.59 


.55 


<< 


*' 


12.25 


3.60 


.32 


2.20 


24.2 


6.9 


2.59 


1.19 


.71 


.57 


.53 


" 


" 


14.75 


4.34 


.42 


2.30 


27.2 


7.8 


2.50 


1.40 


.79 


.57 


.53 


•' 


" 


17.25 


5.07 


.53 


2.41 


30.2 


8.6 


2.44 


1.62 


.87 


.56 


.55 


" 


" 


19.75 


5.81 


.63 


2.51 


33.2 


9.5 


2.39 


1.85 


.96 


.56 


.58 


C25 


8 


11.25 


3.35 


.22 


2.26 


32.3 


8.1 


3.10 


1.33 


.79 


.63 


.58 


" 


" 


13.75 


4.04 


.31 


2.35 


36.0 


9.0 


2.98 


1.55 


.87 


.62 


.56 


" 


" 


16.25 


4.78 


.40 


2.44 


39.9 


10.0 


2.89 


1.78 


.95 


.61 


.56 


" 


" 


18.75 


5.51 


.49 


2.53 


43.8 


11.0 


2.82 


2.01 


1.02 


.60 


.57 


11 


" 


21.25 


6.25 


.58 


2.62 


47.8 


11.9 


2.76 


2.25 


1.11 


.60 


.59 


C29 


9 


13.25 


3.89 


.23 


2.43 


47.3 


10.5 


3.49 


1.77 


.97 


.67 


.61 


" 


" 


15.00 


4.41 


.29 


2.49 


50.9 


11.3 


3.40 


1.95 


1.03 


.66 


.59 


" 


" 


20.00 


5.88 


.45 


2.65 


60.8 


13.5 


3.21 


2.45 


1.19 


.65 


.58 


" 


" 


25.00 


7.35 


.61 


2.81 


70.7 


15.7 


3.10 


2.98 


1.36 


.64 


.62 


C33 


10 


15.00 


4.46 


.24 


2.60 


66.9 


13.4 


3.87 


2.30 


1.17 


.72 


.64 


" 


" 


20.00 


5.88 


.38 


2.74 


78.7 


15.7 


3.66 


2.85 


1.34 


.70 


.61 


" 


" 


25.00 


7.35 


.53 


2.89 


91.0 


18.2 


3.52 


3.40 


1.50 


.68 


.62 


•' 


" 


30.00 


8.82 


.68 


3.04 


103.2 


20.6 


3.42 


3.99 


1.67 


.67 


.65 


" 


" 


35.00 


10.29 


.82 


3.18 


115.5 


23.1 


3.35 


4.66 


1.87 


.67 


.69 


C41 


12 


20.50 


6.03 


.28 


2.94 


128.1 


21.4 


4.61 


3.91 


1.75 


.81 


.70 




" 


25.00 


7.35 


.39 


3.05 


144.0 


24.0 


4.43 


4.53 


1.91 


.78 


.68 


♦• 


" 


30.00 


8.82 


.51 


3.17 


161.6 


26.9 


4.28 


5.21 


2.09 


.77 


.68 


•1 


" 


35.00 


10.29 


.64 


3.30 


179.3 


29.9 


4.17 


5.90 


2.27 


.76 


.69 


41 


" 


.40.00 


11.76 


.76 


3.42 


196.9 


32.8 


4.09 


6.63 


2.46 


.75 


.72 


C53 


15 


33 . 00 


9.90 


.40 


3.40 


312.6 


41.7 


5.62 


8.23 


3.16 


.91 


.79 


" 


" 


35.00 


10.29 


.43 


3.43 


319.9 


42.7 


5.57 


8.48 


3.22 


.91 


.79 


" 


" 


40.00 


11.76 


.52 


3.52 


347.5 


46.3 


5.44 


9.39 


3.43 


.89 


.78 


ai 


•' 


45.00 


13.24 


.62 


3 . 62 


375.1 


.50.0 


5.32 


10.29 


3.63 


.88 


.79 


•• 


" 


50.00 


14.71 


.72 


3.72 


402.7 


53.7 


5.23 


11.22 


3.85 


.87 


.80 




ti 


55.00 


16.18 


.82 


3.82 


430.2 


57.4 


5.16 


12.19 


4.07 


.87 


.82 



JOINT COMMITTEE REPORT— TABLE VIII 213 

Abstract from the Report of the Joint Committee on Concrete 
AND Reinforced Concrete. 

The unit stresses proposed are the safe working stresses for a concrete 
column having a compressive strength of 2000 Ib./sq. in. at 28 days. 
These stresses should be reduced in proportion if a poorer concrete is 
used and may be increased for a stronger concrete. The increase should 
not exceed 25 per cent in any case. 

Columns. 

Plain concrete, length less than 12 diameters 450 Ib./sq. in. 

Reinforced with longitudinal bars only, length less than 

15 diameters 450 Ib./sq. in. 

Reinforced with not less than 1 per cent or more than 4 
per cent of longitudinal bars and spiral bands or 
hoops. 650 Ib./sq. in. 

Beams. 

Maximum compression in concrete 650 Ib./sq. in. 

Diagonal tension or shear without reinforcement 40 Ib./sq. in. 

with ample reinforcement 120 Ib./sq. in. 

Bond stress 80 Ib./sq. in. 

Ratio of moduli of elasticity 15 Ib./sq. in. 

Allowable unit stress in the steel 16,000 Ib./sq. in. 



INDEX 



Axial force, 3. 

Bar, a compound, 169. 

definition of, 3. 

of uniform strength, 22. 
Beams, compound, 171. 

continuous, 109. 

deflection of, 102. 

kinds of, 46. 

maximum stress in, 161. 

of uniform strength, 66. 

overhanging, 65. 

reactions at the supports of, 47. 

reenforced concrete, 178. 

relative strengths of, 64. 

restrained or fixed, 105, 106, 107. 
Bending moment, in beams, 50. 

relation between vertical shear 
and, 63. 
Bond stress in reenforced concrete 

beams, 189. 

Columns, long, 123. 

Euler's formula for, 125. 

parabolic formula for, 136. 

Rankine's formula for, 130. 

Ritter's formula for, 133. 

straight line formula for, 138. 

reenforced columns, 193. 
Compound bars and beams, 169, 171. 
Compression combined with shear, 
154. 

formula for, 3. 
Concrete beams, reenforced, 178. 
Continuous beams, 109. 

Dangerous section in beams, 63. 
Deflection of beams, 102. 
Deflection of reenforced concrete 

beams, 195. 
Deformation of elastic bodies, 7. 

unit, 7. 
Diagonal stress in reenforced con- 
crete beams, 190. 
Ductility, 11. 



Elastic limit, 8. 

commercial, 9. 
Elastic curve, 48, 99. 

equation of the, 99. 
Elasticity, modulus of, 8. 
Euler's formula for long columns, 
123. 

Factors of safety, 14. 
Force, internal and external, 6. 
Formulas, table of fundamental, 203. 
table of derived, 204. 

Helical springs, 93. 

Horizontal shear in beams, 158. 

Load, concentrated or uniform — on 
beams, 47. 
eccentric axial — on beams, 153. 
moving — on beams, 67. 

Materials, constants of, 13, 209. 
Modulus of, 

elasticity, tension or compres- 
sion, 8. 

elasticity, flexure, 102, 

elasticity, torsion, 88. 

rupture, tension or compression, 
3. 

rupture, for beams, 55. 

rupture, for torsion, 92. 

section, beams, 56. 

section, torsion, 86. 

for beams, 56. 
Moment, bending, 50. 

diagram, 58. 

resisting, 52. 

Neutral axis or plane, 53. 

Parabolic formula for long columns, 

136. 
Pipes, thin, strength of, 25. 
thick, strength of, 27. 



215 



216 



INDEX 



Rankine's formula for long columns, ' 

130. 
Reactions for beams, 47. 
Reenforced concrete beams, 178. 
Resilience, 10. 

ultimate, 11. 

elastic, 12. 
Ritter's formula for long columns, 

133. 
Riveted joints, 30. 

compression in, 33. 

efficiency of, 37. 

general case of, 33. 

kinds of, 35. 

shear in, 32. 

tension in, 30. 

Sections, properties of, 210.' 

square — in torsion, 86. 
Shafts, couplings for, 90. 

horse power of, 89. 

strength and stiffness of, 89. 

twist of, 88. 

twisting moment in, 83. 
Shear, and axial stress, 154. 

diagrams, 58, 59. 

horizontal shear in beams, 152. 

resisting, 51. 

vertical — in beams, 49. 
Shearing stress, 5. 

Shearing stress in reenforced con- 
crete beams, 188. 
Springs, helical, 93. 
Straight line formulas for long 

columns, 138. 
Strength, bars of uniform, 22. 

of cylinders, pipes, and spheres, 
25. 

of thick pipes, 27. 

ultimate — of materials, 9. 



Stress, 3. 

Stress, combined, 148, 153, 154. 
in roof rafters, 152. 
in long columns, 131. 
due to change of temperature, 39. 
in reenforced concrete beams, 
bond, 189. 
diagonal, 190. 

horizontal tension and com- 
pression, 179. 
shearing, 188. 
maximum — in beams, 63, 161. 
maximum — tensile or compres- 
sive, 5. 

Tables: allowable stresses in reen- 
forced concrete, 213, 
constants of materials, 209. 
derived fownulas, 204. 
fundamental formulas, 203. 
properties of beams, 208. 
properties of sections, 210. 
properties of standard I-beams, 

211. 
properties of standard channel- 
beams, 212. 
Tension, combined with bending, 
144. 
combined with shear, 149. 
formula for, 3, 178. 
Torsion, derivation of formula for, 
83. 
modulus of elasticity for, 183. 

Vertical shear, 49. 

relation between the bending 
moment and, 63. 

Yield point, 9. 



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Bowie, A. J., Jr. A Practical Treatise on Hydraulic Mining Bvo, 5 00 

Bowker, W. R* Dynamo, Motor and Switchboard Circuits Bvo, *2 50 

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Cathcart, W. L., and Chaffee, J. 1. Elements of Graphic Statics. . Svo, *3 od 

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Chalkley, A. P. Diesel Engines Svo, *3 00 

Chambers' Mathematical Tables Svo, i 75 

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Charpentier, P. Timber Svo, *6 00 



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Cornwall, H. B. Manual of Blow-pipe Analysis 8vo, 

Courtney, C. F. Masonry Dams 8vo, 

Cowell, W. B. Pure Air, Ozone, and Water i2mo, 

Craig, T. Motion of a Solid in a Fuel. (Science Series No. 49.) . i6mo, 

Wave and Vortex Motion. (Science Series No. 43.) i6mo, 

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Creedy, F. Single Phase Commutator Motors 8vo, 

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Crocker, F. B., and Arendt, M. Electric Motors 8vo, *2 50 

Crocker, F. B., and Wheeler, S. S. The Management of Electrical Ma- 
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Crosskey, L. R., and Thaw, J. Advanced Perspective 8vo, i 50 

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Dadourian, H. M. Analytical Mechanics i2mo, *3 00 

Danby, A. Natural Rock Asphalts and Bitumens 8vo, *2 50 

Davenport, C. The Book. (Westminster Series.) 8vo, *2 00 

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DelaCoux, H. The Industrial Uses of Water. Trans, by A. Morris. 8vo, *4 50 

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Draper, C. H. Elementary Text-book of Light, Heat and Sound . . i2mo, 

Heat and the Principles of Thermo-dynamics lamo, 

Dubbel, H. High Power Gag Engines 8vo, 

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Dumesny, P., and Noyer, J. Wood Products, Distillates, and Extracts. 

8vo, 
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Svo, 
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Evans, C. A. Macadamized Roads (In Press.) 



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Fanning, J. T. Hydraulic and Water-supply Engineering 8vo, 

Fauth, P. The Moon in Modern Astronomy. Trans, by J. McCabe. 

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Fernbach, R. L. Glue and Gelatine 8vo, 

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Fisher, H. K. C, and Darby, W. C. Submarine Cable Testing 8vo, 

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Vol. II. The Utilization of Induced Currents *5 00 

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Flynn, P. J. Flow of Water. (Science Series No. 84.) i2mo, o 50 

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Foster, H. A. Electrical Engineers' Pocket-book. (Seventh Edition.) 

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Foster, Gen. J. G. Submarine Blasting in Boston (Mass.) Harbor 4to, 350 

Fowle, F. F. Overhead Transmission Line Crossings i2mo, *i 50 

The Solution of Alternating Current Problems 8vo (In Press.) 

Fox, W. G. Transition Curves. (Science Series No. no.) i6mo, o 50 

Fox, W., and Thomas, C. W. Practical Course in Mechanical Draw- 
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Fritsch, J. Manufacture of Chemical Manures. Trans, by D. Grant. 

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Gant, L. W. Elements of Electric Traction 8vo, *2 50 

Garcia, A. J. R. V. Spanish-English Railway Terms 8vo, *4 50 

Garforth, W. E. Rules for Recovering Coal Mines after Explosions and 

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Gear, H. B., and "Williams, P. F. Electric Central Station Distribution 

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Geikie, J. Structural and Field Geology 8vo, *4 00 

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Goodeve, T. M. Textbook on the Steam-engine i2mo, 2 00 

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Gray, J. Electrical Influence Machin2s i2mo, 

MarSne Boiler Design i2mo, 

Greenhill, G. Dynamics of Mechanical Flight .8vo, 

I Greenwood, E. Classified Guide to Technical and Commercial Books. 8vo, 

Gregorius, R. Mineral Waxes. Trans, by C. Salter i2mo, 

Griffiths, A. B. A Treatise on Manures i2mo, 

Dental Metallurgy 8vo, 

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Giildner, Hugo. Internal Combustion Engines. Trans, by H. Diederichs. 

4to, *io 00 

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Haeder, H. Handbook on the Steam-engine. Trans, by H. H. P. 

Powles i2mo, 

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Hall, C. H. Chemistry of Paints and Paint Vehicles i2mo, 

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Halsey, F. A. Slide Valve Gears i2mo, 

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Worm and Spiral Gearing. (Science Series No. 116.) i6mo, 

Hamilton, W. G. Useful Information for Railway Men i6mo, 

Hammer, W. J. Radium and Other Radio-active Substances 8vo, 

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Hatt, J. A. H. The Colorist square i2mo, *i 50 

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Wright i2mo, *2 co 

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Wright 8vo, *5 00 

Hausner, A. Manufacture of Preserved Foods and Sweetmeats. Trans. 

by A. Morris and H. Robson 8vo, 

Hawke, W. H. Premier Cipher Telegraphic Code 4to, 

■ 100,000 Words Supplement to the Premier Code 4to, 

Hawkesworth, J. Graphical Handbook for Reinforced Concrete Design. 

4to, 

Hay, A. Alternating Currents 8vo, 

Electrical Distributing Networks and Distributing Lines Bvo, 

Continuous Current Engineering Bvo, 

Hayes, H. V. Public Utilities, Their Cost New and Depreciation. . .8vo, 

Heap, Major D. P. Electrical Appliances 8vo, 

Heather, H. J. S. Electrical Engineering 8vo, 

lieaviside, O. Electromagnetic Theory. Vols. I and II. . . .8vo, each. 

Vol. Ill 8vo, 

Heck, R.~ C. H. The Steam Engine and Turbine 8vo, 

Steam-Engine and Other Steam Motors. Two Volumes. 

Vol. I. Thermodynamics and the Mechanic5> 8vo, 

Vol. II. Form, Construction, and Working 8vo, 

■ Notes on Elementary Kinematics. 8vo, boards, 

Graphics of Machine Forces 8vo, boards. 

Hedges, K. Modern Lightning Conductors 8vo, 

Heermann, P. Dyers' Materials. Trans, by A. C. Wright i2mo, 

Hellot, Macquer and D'Apligny. Art of Dyeing Wool, Silk and Cotton. Bvo, 

Henrici, 0. Skeleton Structures 8vo, 

Hering, D. W. Essentials of Physics for College Students Bvo, 

Hering-Shaw, A. Domestic Sanitation and Plumbing. Two Vols.. .Bvo, 

Hering-Shaw, A. Elementary Science Bvo, 

Herrmann, G. The Graphical Statics of Mechanism. Trans, by A. P. 

Smith i2mo, 

Herzfeld, J. Testing of Yarns and Textile Fabrics Bvo, 

Hildebrandt, A. Airships, Past and Present .8vo, 

Hildenbrand, B. W. Cable-Making, (Science Series No. 32.) . .. .i6mo, 
Hilditch, T. P. A Concise History of Chemistry lamo, 

Hill, J. W. The Purification of Public Water Supplies. New Edition. 

(In Press.) 

Interpretation of Water Analysis (in Press.) 

Hill, M. J. M. The Theory of Proportion Bvo, *2 50 

Hiroi, I. Plate Girder Construction. (Science Series No. g5.)...i6mo, o 50 

Statically-Indeterminate Stresses i2mo, *2 00 

Hirshfeld, C. F. Engineering Thermodynamics. (Science Series No. 45,) 

i6mo, o 50 

Hobart, H. M. Heavy Electrical Engineering Bvo, *4 50 

Design of Static Transformers i2mo, *2 00 

Electricity 8vo, *2 00 

Electric Trains ...Bvo, *2 50 



■"3 


00 


*5 


00 


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03 


*2 


50 


*2 


50 


*3 


50 


*2 


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*5 


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50 


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00 


*i 


03 


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00 


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*3 


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50 





50 


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25 



14 D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 

Hobart, H. M. Electric Propulsion of Ships 8vo, *2 oo 

Hobart, J. F. Hard Soldering, Soft Soldering and Brazing i2mo, *i oo 

Hobbs, W. R. P. The Arithmetic of Electrical Measurements i2mo,^ o 50 

Hoff, J. N. Paint and Varnish Facts and Formulas i2mo, *i 50 

Hole, W. The Distribution of Gas 8vo, *7 50 

Holley, A. L. Railway Practice folio, 12 00 

Holmes, A. B. The Electric Light Popularly Explained. ..i2mo, paper, o 50 

Hopkins, N. M. Experimental Electrochemistry 8vo, *3 00 

Model Engines and Small Boats i2mo, i 25 

Hopkinson, J., Shoolbred, J. N., and Day, R. E. Dynamic Electricity. 

(Science Series No. 71.) i6mo, o 50 

Horner, J. Metal Turning i2mo, 1 50 

Practical Ironf ounding Svo, *2 00 

Plating and Boiler Making Svo, 3 00 

Gear Cutting, in Theory and Practice Svo, *3 00 

Houghton, C. E. The Elements of Mechanics of Materials i2mo, *2 00 

Houllevigue, L. The Evolution of the Sciences Svo, *2 00 

Houstoun, R. A. Studies in Light Production i2mo, 2 00 

Hovenden, F. Practical Mathematics for Young Engineers i2mo, *i 00 

Howe, G. Mathematics for the Practical Man i2mo, *i 25 

Howorth, J. Repairing and Riveting Glass, China and Earthenware. 

Svo, paper, *o 50 

Hubbard, E. The Utilization of Wood-waste Svo, *2 50 

Hiibner, J. Bleaching and Dyeing of Vegetable and Fibrous Materials. 

(Outlines of Industrial Chemistry.) Svo, *5 00 

Hudson, 0. F. Iron and Steel. (Outlines of Industrial Chemistry.). Svo, *2 00 

Mumper, W. Calculation of Strains in Girders i2mo, 2 50 

Humphrey, J. C. W. Metallography of Strain. (Metallurgy Series.) 

{In Press.) 

Humphreys, A. C. The Business Features of Engineering Practice.. Svo, *i 25 

Hunter, A. Bridge Work Svo. (In Press.) 

Hurst, G. H. Handbook of the Theory of Color Svo, *2 50 

Dictionary of Chemicals and Raw Products Svo, *3 00 

Lubricating Oils, Fats and Greases Svo, *4 00 

3oaps Svo, *5 00 

Hurst, G. H., and Simmons, W. H. Textile Soaps and Oils Svo, *2 50 

Hurst, H. E., and Lattey, R. T. Text-book of Physics Svo, *3 00 

Also published in three parts. 

Part I. Dynamics and Heat *i 25 

Part II. Sound and Light *i 25 

Part III. Magnetism and Electricity *i 50 

Hutchinson, R. W., Jr. Long Distance Electric Power Transmission. 

i2mo, *3 00 
Hutchinson, R. W., Jr., and Thomas, W. A. Electricity in Mining. i2mo, 

(In Press.) 

Hutchinson, W. B. Patents and How to Make Money Out of Them. 

i2mo, I 25 

Hutton, W. S. Steam-boiler Construction Svo, 6 00 

Practical Engineer's Handbook Svo, 7 00 

— — The Works' Manager's Handbook Svo, 6 co 






50 


*2 


00 





5t> 


*2 


50 


*3 


00 


*2 


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50 


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D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 15 

Hyde, E. W. Skew Arches. (Science Series No. 15.) i6mo, 

Hyde, F. S. Solvents, Oils, Gums, Waxes 8vo, 

Induction Coils. (Science Series No. 53.) i6mo, 

Ingham, A. E. Gearing. A practical treatise 8vo, 

Ingle, H. Manual of Agricultural Chemistry Svo, 

Inness, C. H. Problems in Machine Design i2mo, 

Air Compressors and Blowing Engines i2mo, 

Centrifugal Pumps i2mo, 

The Fan i2mo, 

Isherwood, B. F. Engineering Precedents for Steam Machinery . , . 8vo, 
Ivatts, E. B. Railway Management at Stations 8vo, 

Jacob, A., and Gould, E. S. On the Designing and Construction of 

Storage Reservoirs. (Science Series No. 6) i6mo, o 53 

Jannettaz, E. Guide to the Determination of Rocks. Trans, by G. W. 

Plympton i2nio, 

Jehl, F. Manufacture of Carbons 8vo, 

Jennings, A. S. Commercial Paints and Painting. (Westminster Series.) 

8vo, 

Jennison, F. H. The Manufacture of Lake Pigments Svo, 

Jepson, G. Cams and the Principles of their Construction Svo, 

Mechanical Drawing Svo {In Preparation.) 

Jockin, W. Arithmetic of the Gold and Silversmith i2mo, *i 00 

Johnson, J. H. Arc Lamps and Accessory Apparatus. (Installation 

Manuals Series.) i2mo, *o 73 

Johnson, T. M. Ship Wiring and Fitting. (Installation Manuals Series.) 

i2mo, *o 75 
Johnson, W. H. The Cultivation and Preparation of Para Rubber . . Svo, *3 co 

Johnson, W. McA. The Metallurgy of Nickel (In Preparation.) 

Johnston, J. F. W., and Cameron, C. Elements of Agricultural Chemistry 

and Geology i2mo, 2 

Joly, J. Radioactivity and Geology i2mo, -^3 

Jones, H. C. Electrical Nature of Matter and Radioactivity i2mo, *2 

New Era in Chemistry i2mo, *2 

Jones, M. W. Testing Raw Materials Used in Paint i2mo, *2 

Jones, L., and Scard, F. I. Manufacture of Cane Sugar Svo, *5 

Jordan, L. C. Practical Railway Spiral i2mo, leather, *i 

Joynson, F. H. Designing and Construction of Machine Gearing . . Svo, 2 

Jiiptner, H. F. V. Siderology : The Science of Iron Svo, *5 

Kansas City Bridge 4to, 6 00 

Kapp, G. Alternate Current Machinery. (Science Series No. 96.). i6mo, o 50 

Keim, A. W. Prevention of Dampness in Buildings Svo, *2 00 

Keller, S. S. Mathematics for Engineering Students. i2mo, half leather. 

Algebra and Trigonometry, with a Chapter on Vectors *i 75 

Special Algebra Edition *i . 00 

Plane and Solid Geometry *i . 25 

Analytical Geometry and Calculus *2 00 

Kelsey, W. R. Continuous-current Dynamos and Motors Svo, *2 50 

Kemble, W. T., and Underbill, C. R. The Periodic Law and the Hydrogen 

Spectrum Svo, paper, *o 50 



l6 D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 

Kemp, J. F. Handbook of Rocks 8vo, *i 50 

Kendall, E. Twelve Figure Cipher Code 4to, *i2 50 

Kennedy, A. B. W., and Thurston, R. H. Kinematics of Machinery. ^ 

(Science Series No. 54.) i6mo, o 50 

Kennedy, A. B. W., Unwin, W. C, and Idell, F. E. Compressed Air. 

(Science Series No. 106.) i6mo, o 50 

Kennedy, R. Modern Engines and Power Generators. Six Volumes. 4to, 15 00 

Single Volumes each, 3 00 

■ Electrical Installations. Five Volumes ; . 4to, 15 00 

Single Volumes each, 3 50 

Flying Machines; Practice and Design i2mo, *2 00 

Principles of Aeroplane Construction ^ 8vo, *i 50 

Kennelly, A. E. Electro-dynamic Machinery 8vo, i 50 

Kent, W. Strength of Materials. (Science Series No. 41.) i6mo, o 50 

Kershaw, J. B. C. Fuel, Water and Gas Analysis Svo, *2 50 

• Electrometallurgy. (Westminster Series.) Svo, *2 00 

The Electric Furnace, in Iron and Steel Production i2mo, *i 50 

Electro-Thermal Methods of Iron and Steel Production. .. .8vo, *3 00 

Kinzbrunner, C. Alternate Current Windings Svo, *i 50 

Continuous Current Armatures , 870, *i 50 

Testing of Alternating Current Machines 8vo, *2 00 

Kirkaldy, W. G. David Kirkaldy's System of Mechanical Testing. .4to, 10 00 

Kirkbride, J. Engraving for Illustration 8/0, *i 50 

Eirkwood, J. P. Filtration of River Waters 4to, 7 50 

Kirschke, A. Gas and Oil Engines i2mo, *i 25 

Klein, J. F. Design of a High-speed Steam-engine 8vo, *5 00 

Physical Significance of Entropy 8vo, *i 50 

Kleinhans, F. B. Boiler Construction 8vo, 3 00 

Knight, R.-Adm. A. M. Modern Seamanship 8vo, *7 so 

Half morocco , *9 00 

Knox, J. Physico-Chemical Calculations i2mo, *i 00 

— — Fixation of Atmospheric Nitrogen. (Chemical Monographs, 

No. 4.) lamo, *o 75 

Knox, W. F. Logarithm Tables {In Preparation.) 

Knott, C. G., and Mackay, J. S. Practical Mathematics 8vo, 2 00 

Koester, F. Steam-Electric Power Plants 4to, *5 00 

Hydroelectric Developments and Engineering 4to, *5 00 

Koller, T. The Utilization of Waste Products Svo, *3 50 

■ Cosmetics 8vo, *2 5a 

Kremann, R. Application of the Physico-Chemical Theory to Tech- 
nical Processes and Manufacturing Methods. Trans, by H. 

E. Potts Svo, *2 50 

Kretchm^r, K. Yarn and Warp Sizing 8vo, *4 oo 

Lallier, E. V. Elementary Manual of the Steam Engine i2mo, *2 00 

Lambert, T. Lead and Its Compounds 8vo, *3 50 

Bone Products and Manures 8vo, *3 00 

Lamborn, L. L. Co.ttonseed Products Svo, *3 00 

Modern Soaps, Candles, and Glycerin Svo, *7 50 

Lamprecht, R. Recovery Work After Pit Fires. Trans, by C. Salter . Svo, *4 oa 

Lancaster, M. Electric Heating, Cooking and Cleaning Svo, *i 50 



D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 17 

Lanchester, F. W. Aerial Flight. Two Volumes. 8vo. 

Vol. I. Aerodynamics *6 00 

Aerial Flight. Vol. II. Aerodonetics *6 . 00 

Lamer, E. T. Principles of Alternating Currents i2mo. *i 25 

Larrabee, C. S. Cipher and Secret Letter and Telegraphic Code. i6mo, o 60 

La Rue, B. F. Swing Bridges. (Science Series No. 107.) i6mo, o 50 

Lassar-Cohn. Dr. Modern Scientific Chemistry. Trans, by M. M. 

Pattison Muir i2mo, *2 00 

Latimer, L. H., Field, C. J., and Howell, J. W. Incandescent Electric 

Lighting. (Science Series No. 57.) i6mo, o 50 

Latta, M. N. Handbook of American Gas-Engineering Practice . . . 8vo, *4 50 

American Producer Gas Practice 4to, *6 00 

Laws, B. C. Stability and Equilibrium of Floating Bodies 8vo, *3 50 

Lawson, W. R. British Railways. A Financial and Commercial 

Survey 8yo, 200 

Leask, A. R. Breakdowns at Sea i2mo, 2 00 

Refrigerating Machinery i2mo, 2 00 

Lecky, S. T. S. " Wrinkles " in Practical Navigation 8vo, *8 00 

Le Doux, M. Ice-Making Machines. (Science Series No. 46.) .. i6mo, 050 

Leeds, C. C. Mechanical Drawing for Trade Schools oblong 4to, 

High School Edition *i 25 

Machinery Trades Edition *2 . 06 

Lefevre, L. Architectural Pottery. Trans, by H. K. Bird and W. M. 

Binns 4to, *7 50 

Lehner, S. Ink Manufacture. Trans, by A. Morris and H. Robson . 8vo, *2 50 

Lemstrom, S. Electricity in Agriculture and Horticulture 8vo, *i 50 

Letts, E. A. Fundamental Problems in Chemistry Bvo, *2 00 

Le Van, W; B. Steam-Engine Indicator. (Science Series No. 78.)i6mo, o 50 
Lewes, V. B. Liquid and Gaseous Fuels. (Westminster Series.) . .8vo, *2 00 

Carbonization of Coal 8vo, *3 00 

Lewis, L. P. Railway Signal Engineering 8vo, *3 50 

Lieber, B. F. Lieber's Standard Telegraphic Code Bvo, *io 00 

Code. German Edition 8vo, *io 00 

Spanish Edition 8vo, *io 00 

French Edition 8vo, *io 00 

Terminal Index 8vo, *2 50 

Lieber's Appendix folio, *i5 00 

Handy Tables 4to, *2 50 

Bankers and Stockbrokers' Code and Merchants and Shippers' 

Blank Tables 8vo, *i5 00 

100,000,000 Combination Code 8vo, *io 00 

Engineering Code 8vo, *i2 50 

Livermore, V. P., and Williams, J. How to Become a Competent Motor- 
man i2mo, *i 00 

Liversedge, A. J. Commercial Engineering 8vo, *3 00 

Livingstone, R. Design and Construction of Commutators 8vo, *2 25 

Mechanical Design and Construction of Generators Bvo, *3 50 

Lobben, P. Machinists' and Draftsmen's Handbook 8vo, 2 50 

Lockwood, T. D. Electricity, Magnetism, and Electro-telegraph .... 8vo, 2 50 



I8 D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 

Lockwood, T. D. Electrical Measurement and the Galvanometer. 

z2mo, ^ 75 

Lodge, O. J. Elementary Mechanics i2mo, i 50 

Signalling Across Space without Wires 8vo, *2 00 

Loewenstein, L. C, and Crissey, C. P. Centrifugal Pumps *4 5© 

Lord, R. T. Decorative and Fancy Fabrics 8vo, *3 50 

Loring, A. E. A Handbook of the Electromagnetic Telegraph .... i6mo o 50 

— — Handbook. (Science Series No. 39.) i6mo, o 50 

Xow, D. A. Applied Mechanics (Elementary) i6mo, o 80 

Lubschez, B. J. Perspective i2mo, *i 50 

Xucke, C. E. Gas Engine Design 8vo, *3 00 

• Power Plants: Design, EflBlciency, and Power Costs. 2 vols. 

(In Preparation.) 

Xunge, G. Coal-tar and Ammonia. Two Volumes 8vo, *i5 00 

■ Manufacture of Sulphuric Acid and Alkali. Four Volumes .... 8vo, 

Vol. I. Sulphuric Acid. In three parts *i 8 co 

Vol. n. Salt Cake, Hydrochloric Acid and Leblanc Soda. In two 

parts *i5 . 00 

Vol. III. Ammonia Soda *io 00 

Vol. IV. Electrolytic Methods (In Press.) 

Technical Chemists' Handbook i2mo, leather, *3 50 

Technical Methods of Chemical Analysis. Trans, by C. A. Keane 

in collaboration with a corps of specialists. Three Volumes. ^48 00 

Vol. I. In two parts 8vo, *i5 00 

Vol. II. In two parts 8vo, *i8 00 

Vol. III. In two parts 8vo, *i8 00 

Lupton, A., Parr, G. D. A., and Perkin, H. Electricity as Applied to 

Mining 8vo, *4 50 

Luquer, L. M. Minerals in Rock Sections 8vo, *i 50 

J-Iacewen, H. A. Food Inspection 8vo, *2 50 

Mackenzie, N. F. Notes on Irrigation Works 8vo, *2 50 

Mackie, J. How to Make a Woolen Mill Pay 8vo, *2 00 

Mackrow, C. Naval Architect's and Shipbuilder's Pocket-book. 

i6mo, leather, 5 00 

Maguire, Wm. R. Domestic Sanitary Drainage and Plumbing . . . .8vo, 4 00 
Mallet, A. Compound Engines. Trans, by R. R. Buel. (Science Series 

No. 10.) i6mo, 

Mansfield, A. N. Electro-magnets. (Science Series No. 64.) . . . i6mo, o 50 

Marks, E. C. R. Construction of Cranes and Lifting Machinery .i2mo, *i 50 

Construction and Working of Pumps i2mo, *i 50 

• Manufacture of Iron and Steel Tubes i2mo, *2 00 

■ Mechanical Engineering Materials i2mo, *i 00 

Marks, G. C. Hydraulic Power Engineering Svo, 3 50 

Inventions, Patents and Designs i2mo, *i 00 

Marlow, T. G. Drying Machinery and Practice Svo, *5 00 

Marsh, C. F. Concise Treatise on Reinforced Concrete Svo, *2 50 

• Reinforced Concrete Compression Member Diagram. Mounted on 

Cloth Boards *i . 50 



D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 19 

Marsh, C. F., and Dunn, W. Manual of Reinforced Concrete and Con- 
crete Block Construction i6mo, morocco, *2 50 

Marshall, W. J., and Sankey, H. R. Gas Engines. (Westminster Series.) 

8vo, *2 00 

Martin, G. Triumphs and Wonders of Modem Chemistry 8vo, *2 00 

Martin, N. Properties and Design of Reinforced Concrete i2mo, *2 50 

jyiartin, W. D. Hints to Engineers i2mo, *i 00 

Massie, W. W., and Underhill, C. R. Wireless Telegraphy and Telephony. 

i2mo, *i 00 
Matheson, D. Australian Saw-Miller's Log and Timber Ready Reckoner. 

i2mo, leather, i 50 

Mathot, R. E. Internal Combustion Engines 8vo, *6 00 

Maurice, W. Electric Blasting Apparatus and Explosives . Svo, *3 50 

Shot Firer's Guide Svo, *i 50 

Maxwell, J. C. Matter and Motion. (Science Series No. 36.). 

i6mc, o 50 
Maxwell, W. H., and Brown, J. T. Encyclopedia of Municipal and Sani- 
tary Engineering 4to, *io 00 

Mayer, A. M. Lecture Notes on Physics Svo, 2 00 

McCuUough, R. S. Mechanical Theory of Heat Svo, 3 50 

McGibbon, W. C. Indicator Diagrams for Marine Engineers Svo, *3 00 

Marine Engineers' Drawing Book oblong 4to, *2 00 

Mcintosh, J. G. Technology of Sugar Svo, *4 50 

Industrial Alcohol Svo, *3 00 

■ Manufacture of Varnishes and Kindred Industries. Three Volumes. 

Svo. 

Vol. I. Oil Crushing, Refining and Boiling *3 50 

Vol. ' II. Varnish Materials and Oil Varnish Making *4 00 

Vol. III. Spirit Varnishes and Materials *4 50 

McKnight, J. D., and Brown, A. W. Marine Multitubular Boilers *i 50 

McMaster, J. B. Bridge and Tunnel Centres. (Science Series No. 20.) 

i6mo, o 50 

McMechen, F. L. Tests for Ores, Minerals and Metals i2mo, *i 00 

McPherson, J. A. Water-works Distribution Svo, 2 50 

Melick, C. W. Dairy Laboratory Guide i2mo, *i 25 

Merck, E. Chemical Reagents ; Their Purity and Tests. Trans, by 

H. E. Schenck Svo, i 00 

JVIerivale, J. H. Notes and Formulae for Mining Students lamo, i 50 

Merritt, Wm. H. Field Testing for Gold and Silver i6mo, leather, i 50 

Messer, W. A. Railway Permanent Way Svo {In Press.) 

Meyer, J. G. A., and Pecker, C. G. Mechanical Drawing and Machine 

Design 4to, 5 oo 

Michell, S. Mine Drainage Svo, 10 00 

Mierzinski, S. Waterproofing of Fabrics. Trans, by A. Morris and H. 

Robson Svo, *2 50 

Miller, G. A. Determinants. (Science Series No 105.) i6mo, 

Miiroy, M. E. W. Home Lace-making i2mo, *i 00 

Minifie, W. Mechanical Drawing Svo, *4 00 

Mitchell, C. A. Mineral and Aerated Waters. , Svo, *3 00 



20 D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 

Mitchell, C. A., and Prideaux, R. M. Fibres Used in Textile and Allied 

Industries 8vo, *3 a,© 

Mitchell, C. F., and G. A. Building Construction and Drawing. i2mo. 

Elementary Course *i 50 

Advanced Course *2 50 

Monckton, C. C. F. Radiotelegraphy. (Westminster Series.) 8vo, *2 00 

Monteverde, R. D. Vest Pocket Glossary of English-Spanish, Spanish- 
English Technical Terms 64mo, leather, *i 00 

Montgomery, J. H. Electric Wiring Specifications {In Press.) 

Moore, E. C. S, New Tables for the Complete Solution of Ganguillet and 

Kutter's Formula 8vo, *5 00 

Morecroft, J. H., and Hehre, F. W. Short Course in Electrical Testing. 

Svo, *i 50 
Moreing, C. A., and Neal, T. New General and Mining Telegraph Code. 

8vo, *5 00 

Morgan, A. P. Wireless Telegraph Apparatus for Amateurs i2mo, *i 50 

Moses, A. J. The Characters of Crystals , 8vo, *2 00 

and Parsons, C. L. Elements of Mineralogy Svo, *2 50 

Moss, S.A. Elementsof Gas Engine Design. (Science Series No. 121.) i6mo, o 50 

The Lay-out of Corliss Valve Gears. (Science Series No. ii9.)i6mo, o 50 

Mulford, A. C. Boundaries and Landmarks i2mo, *i 00 

Mullin, J. P. Modern Moulding and Pattern-making i2mo, 2 50 

Munby, A. E. Chemistry and Physics of Building Materials. (West- 
minster Series.) 8vo, *2 00 

Murphy, J. G. Practical Mining i6mo, i 00 

Murphy, W. S. Textile Industries. Eight Volumes *2o 00 

Sold separately, each, '^3 00 

Murray, J. A. Soils and Manures. (Westminster Series.) 8vo, *2 00 

Naquet, A. Legal Chemistry i2mo, 2 00 

Nasmith, J. The Student's Cotton Spinning Svo, 3 00 

Recent Cotton Mill Construction i2mo, 2 00 

Neave, G. B., and Heilbron, I. M. Identification of Organic Compounds. 

i2mo, *i 25 

Neilson, R. M. Aeroplane Patents 8vo, *2 00 

Nerz, F. Searchlights. Trans, by C. Rodgers 8vo, *3 00 

Neuberger, H., and Noalhat, H. Technology of Petroleum. Trans, by 

J. G. MoLitosh 8vo, *io 00 

Newall, J. W. Drawing, Sizing and Cutting Bevel-gears Svo, i 50 

Nicol, G. Ship Construction and Calculations Svo, *4 50 

Nipher, F. E. Theory of Magnetic Measuremrents i2mo, i 00 

Nisbet, H. Grammar of Textile Design Svo, *3 00 

Nolan, H. The Telescope. (Science Series No. 51.) i6mo, o 50 

Noll, A. How to Wire Buildings i2mo, i 50 

North, H. B. Laboratory Experiments in General Chemistry i2mo, *i co 

Nugent, E. Treatise on Optics i2mo, i 50 

O'Connor, H. The Gas Engineer's Pocketbook i2mo, leather, 3 50 

Petrol Air Gas i^mo, *o 75 



D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 21 

Ohm, G. S., and Lockwood, T. D. Galvanic Circuit. Translated by 

William Francis. (Science Series No. 102.) i6mo, o 50 

Olsen, J. C. Text-book of Quantitative Chemical Analysis 8vo, *4 00 

Olsson, A. Motor Control, in Turret Turning and Gun Elevating. (U. S. 

Navy Electrical Series, No. i.) i2mo, paper, *o 50 

Ormsby, M. T. M. Surveying lamo, i 50 

Cudin, M. A. Standard Polyphase Apparatus and Systems 8vo, *3 00 

[)wen, D. Recent Physical Research 8vo, *i 50 

Pakes, W. C. C, and Nankivell, A. T. The Science of Hygiene . .8vo, *i 75 

Palaz, A. Industrial Photometry. Trans, by G. W. Patterson, Jr .. Svo, *4 00 

Pamely, C. Colliery Manager's Handbook Svo, *io co 

Parker, P. A. M. The Control of Water Svo, ^5 00 

Parr, G. D. A. Electrical Engineering Measuring Instruments. .. .8vo, *3 50 

Parry, E. J. Chemistry of Essential Oils and Artificial Perfumes. . .Svo, *5 00 

Foods and Drugs. Tv/o Volumes Svo, 

Vol. I. Chemical and Microscopical Analysis of Foods and Drugs. *7 5° 

Vol. II. Sale of Food and Drugs Act *3 00 

and Coste, J. H. Chemistry of Pigments Svo, *4 50 

Parry, L. A. Risk and Dangers of Various Occupations Svo, *3 00 

Parshall, H. F., and Hobart, H. M. Armature Windings 4to, *7 50 

■ Electric Railway Engineering 4to, *io 00 

and Parry, E. Electrical Equipment of Tramways. . (/w Press.) 

Parsons, S. J. Malleable Cast Iron Svo, *2 50 

Partington, J. R. Higher Mathematics for Chemical Students. .i2mo, *2 00 
Textbook of Thermodynamics Svo, *4 00 

Passmore, A. C. Technical Terms Used in Architecture Svo, *3 50 

Patchell, W. H. Electric Power in Mines , . . . . Svo, *4 00 

Paterson, G. W. L. Wiring Calculations i2mo, *2 00 

Patterson, D. The Color Printing of Carpet Yarns Svo, *3 50 

Color Matching on Textiles Svo, *3 00 

The Science of Color Mixing Svo, *3 00 

Paulding, C. P. Condensation of Steam in Covered and Bare Pipes. .Svo, *2 00 

■ Transmission of Heat through Cold-storage Insulation i2mo, *i 00 

Payne, D. W. Iron Founders' Handbook (In Press.) 

Peddie, R. A. Engineering and Metallurgical Books i2mo, *i 50 

Peirce, B. System of Analytic Mechanics 4to, 10 00 

Pendred, V. The Railway Locomotive. (Westminster Series.) Svo, *2 00 

Perkin, F. M. Practical Methods of Inorganic Chemistry i2mo, '''i 00 

Perrigo, 0. E. Change Gear Devices Svo, i 00 

Perrine, F. A. C. Conductors for Electrical Distribution Svo, *3 50 

Perry, J. Applied Mechanics Svo, *2 50 

Petit, G, White Lead and Zinc White Paints Svo, *i 50 

Petit, R. How to Build an Aeroplane. Trans, by T. O'B. Hubbard, and 

J. H. Ledeboer Svo, *i 50 

Pettit, Lieut. J. S. Graphic Processes. (Science Series No. 76.) ... i6mo, o 53 
Philbrick, P. H. Beams and Girders. (Science Series No. S8.) . . . i6mo, 

Phillips, J. Engineering Chemistry Svo, *4 50 

Gold Assaying Svo, *2 50 

— — Dangerous Goods , Svo, 3 50 



22 D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 

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Logarithms for Beginners i2mo. boards, 

The Slide Rule i2mo, 

Plattner's Manual of Blow-pipe Analysis. Eighth Edition, revised. Trans. 

by H. B. Cornwall 8vo, 

Plympton, G. W. The Aneroid Barometer. (Science Series No. 35.) i6mo, 

How to become an Engineer. (Science Series No. 100.) i6mo, 

Van Nostrand's Table Book. (Science Series No. 104.) i6mo, 

Pochet, M. L. Steam Injectors. Translated from the French. (Science 

Series No. 29.) i6mo, 

Pocket Logarithms to Four Places. (Science Series No. 65.) i6mo, 

leather, 

PoUeyn, F. Dressings and Finishings for Textile Fabrics 8vo, 

Pope, F. G. Organic Chemistry i2mo, 

Pope, F. L. Modern Practice of the Electric Telegraph 8vo, 

Popplewell, W. C. Elementary Treatise on Heat and Heat Engines. . i2mo, 

— — Prevention of Smoke Svo, 

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No. 3.) i2mo, 

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Potter, T. Concrete Svo, 

Potts, H. E. Chemistry of the Rubber Industry. (Outlines of Indus- 
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Pray, T., Jr. Twenty Years with the Indicator Svo, 

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Prelini, C. Earth and Rock Excavation Svo, 

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Prescott, A. B. Organic Analysis Svo, 

Prescott, A. B., and Johnson, 0. C. Qualitative Chemical Analysis. . . Svo, 
Prescott, A. B., and Sullivan, E. C. First Book in Qualitative Chemistry. 

i2mo, 

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Primrose, G. S. C. Zinc. (Metallurgy Series.) (In Press.) 

Pritchard, 0. G. The Manufacture of Electric-light Carbons . . Svo, paper, *o 60 
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Pulsifer, W. H. Notes for a History of Lead Svo, 4 00 

Purchase, W. R. Masonry i2mo, *3 00 

Putsch, A. Gas and Coal-dust Firing Svo, *3 00 

Pynchon, T. R. Introduction to Chemical Physics Svo, 3 00 



*I 


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D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 23 

Rafter G. W. Mechanics of Ventilation. (Science Series No. 33.) . i6mo, 

Potable Water. (Science Series No. 103.) i6mo, 

Treatment of Septic Sewage. (Science Series No. 118.) , , .i6mo, 

Rafter, G. W., and Baker, M. N. Sewage Disposal in the United States. 

4to, 

Raikes, H. P. Sewage Disposal Works 8vo, 

Randall, P. M. Quartz Operator's Handbook 12010, 

Randau, P. Enamels and Enamelling , 8vo, 

Rankine, W. J. M. Applied Mechanics 8vo, 

■ Civil Engineering , , Svo, 

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--^ — Useful Rules and Tables Svo, 

Rankine, W. J. M., and Bamber, E. F. A Mechanical Text-book Svo, 

Raphael, F. C. Localization of Faults in Electric Light and Power Mains. 

Svo, 

Rasch, E. Electric Arc Phenomena. Trans, by K. Tornberg Svo, 

Rathbone, R. L. B. Simple Jewellery Svo, 

Rateau, A. Flow of Steam through Nozzles and Orifices. Trans, by H. 

B, Brydon Svo 

Rausenberger, F. The Theory of the Recoil of Guns Svo, 

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Rautenstrauch, W., and Williams, J. T. Machine Drafting and Empirical 

Design. 

Part I. Machine Drafting Svo, *i 25 

Part II. Empirical Design {In Preparation.) 

Raymond, E. B. Alternating Current Engineering i2mo, *2 50 

Rayner, H. Silk Throwing and Waste Silk Spinning Svo, *2 50 

Recipes for the Color, Paint, Varnish, Oil, Soap and Drysaltery Trades . Svo, *3 50 

Recipes for Flint Glass Making i2mo, *4 50 

Redfern, J. B., and Savin, J. Bells, Telephones (Installation Manuals 

Series.) i6mo, *o 50 

Redgrove, H. S. Experimental Mensuration i2mo, *i 25 

Redwood, B. Petroleum. (Science Series No. 92.) i6mo, o 50 

Reed, S. Turbines Applied to Marine Propulsion *5 00 

Reed's Engineers' Handbook Svo, *5 00 

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oblong 4to, boards, i 00 

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Reiser, F. Hardening and Tempering of Steel. Trans, by A. Morris and 

H. Robson i2mo, *2 50 

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Renwick, W. G. Marble and Marble Working Svo, 5 00 



24 D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 

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Series No. 99.) i6mo, o 50 

Rhead, G. F. Simple Structural Woodwork i2mo, *i 00 

Rhodes, H. J. Art of Lithography 8vo, ^ 3 50 

Rice, J. M., and Johnson, W. W. A New Method of Obtaining the Differ- 
ential of Functions. i2mo, o 50 

Richards, W. A. Forging of Iron and Steel (In Press.) 

Richards, W. A., and North, H. B. Manual of Cement Testing. . . . i2mo, *i 50 

Richardson, J. The Modern Steam Engine 8vo, *3 50 

Richardson, S. S. Magnetism and Electricity i2mo, *2 00 

Rideal, S. Glue and Glue Testing 8vo, *4 00 

Rimmer, E. J. Boiler Explosions, Collapses and Mishaps 8vo, *i 75 

Rings, F. Concrete in Theory and Practice i2mo, *2 50 

Reinforced Concrete Bridges 4to, *5 00 

Ripper, W. Course of Instruction in Machine Drawing folio, *6 00 

Roberts, F. C. Figure of the Earth. (Science Series No. 79.) ..... i6mo, o 50 

Roberts, J., Jr. Laboratory Work in Electrical Engineering. Svo, *2 00 

Robertson, L. S. Water-tube Boilers Svo, 2 00 

Robinson, J. B. Architectural Composition Svo, *2 50 

Robinson, S. W. Practical Treatise on the Teeth of Wheels. (Science 

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■ — ■ — Railroad Economics. (Science Series No. 59.) i6mo, o 50 

' Wrought Iron Bridge Members. (Science Series No. 60.) i6mo, o 50 

Robson, J. H. Machine Drawing and Sketching Svo, *i 50 

Roebling, J. A. Long and Short Span Railway Bridges folio, 25 00 

Rogers, A. A Laboratory Guide of Industrial Chemistry i2mo, *i 50 

Rogers, A., and Aubert, A. B. Industrial Chemistry Svo, *5 o3 

Rogers, F. Magnetism of Iron Vessels. (Science Series No. 30.) . i6mo, o 5o 
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W. J. Britland and H. E. Potts i2mo, *i 25 

Rollinsj W. Notes on X-Light Svo, *5 00 

RoUinson, C. Alphabets Oblong, i2mo, *i 00 

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' Key to Engines and Engine-running i2mo, 2 50 

Rose, T. K. The Precious Metals. (Westminster Series.) Svo, *2 00 

Rosenhain, W. Glass Manufacture. (Westminster Series.) Svo, *2 00 

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Svo, (In Press.) 

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Roth. Physical Chemistry Svo, *2 00 

Rouillion, L. The Economics of Manual Training Svo, 2 00 

Rowan, F. J. Practical Physics of the Modern Steam-boiler Svo, *3 00 

and Idell, F. E. Boiler Incrustation and Corrosion. (Science 

Series No.' 27.) i6mo, o 50 

Roxburgh, W. General Foundry Practice. (Westminster Series.) .Svo, '^2 00 

Ruhmer, E. Wireless Telephony. Trans, by J. Erskine-Murray . .Svo, *3 50 

Russell, A. Theory of Electric Cables and Networks Svo, *3 00 

Sabine, R. History and Progress of the Electric Telegraph i2mo, i 25 

Saeltzer, A. Treatise on Acoustics i2rro, i 00 



D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 25 

Sanford, P. G. Nitro-explosives 8vo, *4 00 

Satmders, C. H. Handbook of Practical Mechanics i6mo, i 00 

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Saunnier, C. Watchmaker's Handbook i2mo, 3 00 

Sayers, H. M. Brakes for Tram Cars 8vo, *i 25 

Scheele, C. W. Chemical Essays 8vo, *2 00 

Scheithauer, W. Shale Oils and Tars Svo, *3 5o 

Schellen, H. Magneto-electric and Dynamo-electric Machines .... Svo, 5 oj 

Scherer, R. Casein. Trans, by C. Salter Svo, *3 00 

Schidrowitz, P. Rubber, Its Production and Industrial Uses Svo, *5 00 

Schindler, K. Iron and Steel Construction Works i2mo, *i 25 

Schmall, C. N. First Course in Analytic Geometry, Plane and Solid. 

i2mo, half leather, *i 75 

Schmall, C. N., and Shack, S. M. Elements of Plane Geometry. . . i2mo, *i 25 

Schmeer, L. Flow of Water Svo, *3 00 

Schumann, F. A Manual of Heating and Ventilation. .. .i2mo, leather, i 50 

Schwarz, E. H. L. Causal Geology Svo, *2 50 

Schweizer, V. Distillation of Resins .8vo, *3 50 

Scott, W. W. Qualitative Analysis. A Laboratory Manual Svo, *i 50 

Scribner, J. M. Engineers' and Mechanics' Companion. .i6mo, leather, i 50 
Scuddeir, Hj Electrical Conductivity and Ionization Constants of 

Organic Compounds Svo, *3 oo 

Searle, A. B. Modern Brickraaking Svo, *5 00 

Cement, Concrete and Bricks Svo, *3 00 

Searle, G. M. "Sumners' Method." Condensed and Improved. 

(Science Series No. 124.) i6mo, o 50 

Seaton, A. E. Manual of Marine Engineering Svo 8 00 

SeatoU; A. E., and Rounthwaite, H. M. Pocket-book of Marine Engi- 
neering i6mo, leather, 3 50 

Seeligmann, T., Torrilhon, G. L., and Falconnet, H. India Rubber and 

Gutta Percha. Trans, by J. G. Mcintosh Svo, *5 00 

Seidell, A. Solubilities of Inorganic and Organic Substances Svo, *3 00 

Seligman, R. Aluminum. (Metallurgy Series.) {In Press.) 

Sellew, W. H. Steel Rails , 4to, *i2 50 

Senter, G. Outlines of Physical Chemistry i2mo, *i 75 

Text-book of Inorganic Chemistry i2mo, *i 75 

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Sever, G. F., and Tow^nsend, F. Laboratory and Factory Tests in Elec- 
trical Engineering Svo, *2 50 

Sewall, C. H. Wireless Telegraphy Svo, *2 00 

Lessons in Telegraphy i2mo, *i 00 

Sevrell, T. Elements of Electrical Engineering Svo, *3 00 

The Construction of Dynamos Svo, *3 00 

Sexton, A. H. Fuel and Refractory Materials i2mo, *2 50 

■ Chemistry of the Materials of Engineering i2mo, *2 50 

Alloys (Non-Ferrous) Svo, *3 00 

■ The Metallurgy of Iron and Steel Svo, *6 50 

Seymour, A. Practical Lithography 

— — Modern Printing Inks Svo, 



* 



2 00 



26 D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 

Shaw, Henry S. H. Mechanical Integrators. (Science Series No. 83.) 

i6mo, 

Shaw, S. History of the Staffordshire Potteries 8vo, 

• Chemistry of Compounds Used in Porcelain Manufacture. .. .Svo, 

Shaw, W. N. Forecasting Weather Svo, 

Sheldon, S., and Hausmann, E. Direct Current Machines lamo, 

Alternating Current Machines lamo, 

Sheldon, S., and Hausmarnn, E. Electric Traction and Transmission 

Engineering i2mo, 

Sheriif, F. F. Oil Merchants' Manual i2mo. 

Shields, J. E. Notes on Engineering Construction i2mo, 

Shreve, S. H. Strength of Bridges and Roofs Svo, 

Shunk, W. F. The Field Engineer i2mo, morocco, 

Simmons, W. H., and Appleton, H. A. Handbook of Soap Manufacture, 

Svo, 
Simmons, W. H., and Mitchell, C. A. Edible Fats and Oils Svo, 

Simms, F. W. The Principles and Practice of Levelling Svo, 

Practical Tunneling Svo, 

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Simpson, W. Foundations Svo. {In Press.) 

Sinclair, A. Development of the Locomotive Engine. . . Svo, half leather, 
Twentieth Century Locomotive Svo, half leather, 

Sindall, R. W., and Bacon, W. N. The Testing of Wood Pulp Svo, 

Sindall, R. W. Manufacture of Paper. (Westminster Series.) . . . .Svo, 

Sloane, T. O'C. Elementary Electrical Calculations i2mo, 

Smallwood, J. C. Mechanical Laboratory Methods. .. .i2mo, leather. 

Smith, C. A. M. Handbook of Testing, MATERIALS Svo, 

Smith, C. A. M., and Warren, A. G. New Steam Tables Svo, 

Smith, C. F. Practical Alternating Currents and Testing Svo, 

Practical Testing of Dynamos and Motors Svo, 

Smith, F. E. Handbook of General Instruction for Mechanics . . . i2mo, 
Smith, H. G. Minerals and the Microscope 

Smith, J. C. Manufacture of Paint Svo, *3 oc 

Paint and Painting Defects 

Smith, R. H. Principles of Machine Work i2mo, 

Elements of Machine Work i2mo, 

Smith, W. Chemistry of Hat Manufacturing i2mo, 

Snell, A. T. Electric Motive Power Svo, 

Snow, W. G. Pocketbook of Steam Heating and Ventilation. (In Press.) 
Snow, W. G., and Nolan, T. Ventilation of Buildings. (Science Series 

No. 5.) i6mo, 

Soddy, F. Radioactivity Svo, 

Solomon,. M. Electric Lamps. (Westminster Series.) Svo, 

Somerscales, A. N. Mechanics for Marine Engineers i2mo, 

Mechanical and Marine Engineering Science , . .Svo, 

Sothern, J. W. The Marine Steam Turbine Svo, 

• Verbal Notes and Sketches for Marine Engineers Svo, 

Sothern, J. W., and Sothern, R. M. Elementary Mathematics for 

Marine Engineers i2mo, 

Simple Problems in Marine Engineering Design i2mo. 






50 


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D. VAN NOSTRAND CO/S SHORT TITLE CATALOG 27 

Southcombe, J. E. Chemistry of the Oil Industries. (Outlines of In- 
dustrial Chemistry.) 8vo, *3 00 

Soxhlet, D. H. Dyeing and Staining Marble. Trans, by A. Morris and 

H. Robson 8vo, *2 50 

Spang, H. W. A Practical Treatise on Lightning Protection i2mo, i 00 

Spangenburg, L. Fatigue of Metals. Translated by S. H. Shreve. 

(Science Series No. 23.) i6mo, 50 

Specht, G. J., Hardy, A. S., McMaster, J. B., and Walling. Topographical 

Surveying. (Science Series No. 72.) . i6mo, o 50 

Speyers, C. L. Text-book of Physical Chemistry 8vo, *2 25 

Sprague, E. H. Hydraulics i2mo, i 25 

Stahl, A. W. Transmission of Power. (Science Series No. 28.) . i6mo, 

Stahl, A. W., and Woods, A. T. Elementary Mechanism i2mo, *2 00 

Staley, C, and Pierson, G. S. The Separate System of Sewerage. . .8vo, *3 00 

Standage, H. C. Leatherworkers' Manual 8vo, *3 50 

Sealing Waxes, Wafers, and Other Adhesives 8vo, *2 00 

Agglutinants of all Kinds for all Purposes i2mo, *3 50 

Stanley, H. Practical Applied Physics {In Press.) 

Stansbie, J. H. Iron and Steel. (Westminster Series.) 8vo, *2 oo 

Steadman, F. M. Unit Photography , 8vo, *2 00 

Stecher, G, E. Cork. Its Origin and Industrial Uses i2mo, i 00 

Steinman, D. B. Suspension Bridges and Cantilevers. (Science Series 

No. 127.) , o 50 

Stevens, H. P. Paper Mill Chemist i6mo, *2 50 

Stevens, J. S. Theory of Measurements *i 25 

Stevenson, J. L. Blast-Furnace Calculations i2mo, leather, *2 00 

Stewart, A. Modern Polyphase Machinery i2mo, *2 00 

Stewart, G. Modern Steam Traps i2mo, *i 25 

Stiles, A^ Tables for Field Engineers i2mo, i 00 

Stillman, P. Steam-engine Indicator i2mo, i oc 

Stodola, A. Steam Turbines. Trans, by L. C. Loewenstein 8vo, '^5 00 

Stone, H. The Timbers of Commerce 8vo, 3 50 

Stone, Gen. R. New Roads and Road Laws i2mo, i 00 

Stopes, M. Ancient Plants 8vo, *2 00 

— The Study of Plant Life 8vo, *2 00 

Stumpf, Prof. Una-Flow of Steam Engine 4to, *3 50 

Sudborough, J. J., and James, T. C. Practical Organic Chemistry. . i2mo, *2 00 

Suffling, E. R. Treatise on the Art of Glass Painting 8vo, *3 50 

Swan, K. Patents, Designs and Trade Marks. (Westminster Series.). 

8vo, *2 00 
Swinburne, J., Wordingham, C. H., and Martin, T. C. Electric Currents. 

(Science Series No. 109.) i6mo, o 50 

Swoope, C. W. Lessons in Practical Electricity i2mo, *2 00 

Tailfer, L. Bleaching Linen and Cotton Yarn and Fabrics 8vo, *5 00 

Tate, J. S. Surcharged and Different Forms of Retaining-walls. (Science 

Series No. 7.) i6mo, o 50 

Taylor, E. N. Small Water Supplies i2mo, *2 00 

Templeton, W. Practical Mechanic's Workshop Companion. 

i2mo, morocco, 2 00 



28 D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 

Tenney, E. H. Test Methods for Steam Power Plants (In Press.) 

Terry, H.L. India Rubber and its Manufacture. (WestminGter Series.) 

8vo, *2 X>(> 
Thayer, H. R. Structural Design. 8vo. 

Vol. I. Elements of Structural Design *2 oo- 

Vol. II. Design of Simple Structiures *4 oo 

Vol. III. Design of Advanced Structures (In Preparation.) 

Thiess, J. B., and Joy, G. A. Toll Telephone Practice 8vo, *3 5a 

Thorn, C, and Jones, W. H. Telegraphic Connections. .. .oblong, i2mo, i 50 

Thomas, C. W. Paper-makers' Handbook (In Press.) 

Thompson, A. B. Oil Fields of Russia 4to, *7 50 

Petrolevmi Mining and Oil Field Development Svo, *5 00 

Thompson, S. P. Dynamo Electric Machines. (Science Series No. 75.) 

i6mo, o 50 

Thompson, W. P. Handbook of Patent Law of All Countries i6mo, i 5a 

Thomson, G. S. Milk and Cream Testing i2mo, *i 75 

Modern Sanitary Engineering, House Drainage, etc Svo, *3 oa 

Thornley, T. Cotton Combing Machines Svo, *3 00 

Cotton Waste Svo, *3 oa 

Cotton Spinning. Svo. 

First Year *i 50 

Second Year *2 50 

Third Year *2 50 

Thurso, J. W. Modern Turbine Practice Svo, *4 00 

Tidy, C. Meymott. Treatment of Sewage. (Science Series No. 94.)i6mo, o 50 
Tillmans, J. Water Purification and Sewage Disposal. Trans, by 

Hugh S. Taylor Svo, *2 oa 

Tinney, W. H. Gold-mining Machinery Svo, *3 00 

Titherley, A. W. Laboratory Course of Organic Chemistry Svo, *2 00 

Toch, M. Chemistry and Technology of Mixed Paints Svo, *3 00 

Materials for Permanent Painting i2mo, *2 00 

Chemistry and Technology of Mixed Paints, (In two volumes.) 

, (In Press.) 

Tod, J., and McGibbon, W. C. Marine Engineers* Board of Trade 

Examinations Svo, *i 50 

Todd, J., and Whall, W. B. Practical Seamanship Svo, *7 50 

Tonge, J. Coal. (Westminster Series.) Svo, *2 oa 

Townsend, F. Alternating Current Engineering Svo, boards, *o 75 

Townsend, J. Ionization of Gases by Collis'on Svo, *i ^5 

Transactions of the American Institute of Chemical Engineers, Svo. 

Vol. I. 190S *6 00 

Vol. II. 1909 *6 GO 

Vol. III., 1910 *6 00 

Vol. IV. 191 1 *6 00 

Vol. V. 1912 *6 00 

Vol. VI. 1913 *6 00 

Traverse Tables. (Science Series No. 115.) i6mo, 50 

morocco, i 00 

Treiber, E. Foundry Machinery. Trans, by C. Salter icmo, i 25 



D. VAN NOSTRAND CO.'S SHORT TITLE CATALOG 



29 



Trinks, "W., and Housum, C. Shaft Governors. (Science Series No. 122.) 

i6mo, o 50 

Trowbridge, W. P. Turbine Wheels. (Science Series No. 44.) . . i6mo, 50 

Tucker, J. H. A Manual of Sugar Analysis 8vo, 3 50 

Tunner, P. A. Treatise on Roll-turning. Trans, by J. B. Pearse. 

8vo, text and folio atlas, 10 00 
Turnbull, Jr., J., and Robinson, S. W. A Treatise on the Compound 

Steam-engine. (Science Series No. 8.) i6mo, 

Turrill, S. M. Elementary Course in Perspective i2mo, *i 25 

Underbill, C. R. Solenoids, Electromagnets and Electromagnetic Wind- 
ings i2mo, *2 00 

Underwood, N., and Sullivan, T. V. Chemistry and Technology of 

Printing Inks *3 00 

Urquhart, J. W. Electric Light Fitting i2mo, 2 00 

Electro-plating i2mo, 2 00 

Electrotyping i2mo, 2 00 

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Usbome, P. O. G. Design of Simple Steel Bridges 8vo, *4 00 

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D. V^AN NOSTRAND CO.'S SHORT TITLE CATALOG 31 

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